Brilliant Integration Contest - Season 3 (Part 2)

Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 3(Part 1)

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

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#Calculus

Note by Aditya Kumar
4 years, 5 months ago

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Problem 26 :

Show that 0ln(as+xsbs+xs)dx  =  π(ab)cosecπsa,b>0,s>1  . \int_0^\infty \ln\left(\frac{a^s + x^s}{b^s + x^s}\right)\,dx \; =\; \pi(a-b)\mathrm{cosec}\tfrac{\pi}{s} \hspace{1cm} a,b > 0\,,\,s > 1 \;.

This problem has been solved by Fdp Dpf.

Mark Hennings - 4 years, 5 months ago

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Solution problem 26:

Let , F(a)=0+ln(as+xsbs+xs)dx\displaystyle F(a)=\int_0^{+\infty} \ln\left(\dfrac{a^s+x^s}{b^s+x^s}\right)dx

Observe that, F(b)=0F(b)=0

F(a)=sas10+1as+xsdx\displaystyle F^\prime(a)=sa^{s-1}\int_0^{+\infty} \dfrac{1}{a^s+x^s}dx

Perform the change of variable,

y=xay=\dfrac{x}{a},

F(a)=s0+11+xsdx\displaystyle F^\prime(a)=s\int_0^{+\infty} \dfrac{1}{1+x^s}dx

It's well known that,

0+11+xsdx=πscosec(πs)\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)

Therefore,

F(a)=πcosec(πs)\displaystyle F^\prime(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)

Therefore, F(a)=πcosec(πs)a+kF(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)a+k ,

k a real constant.

Since F(b)=0F(b)=0 then k=πcosec(πs)bk=-\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)b

Therefore,

F(a)=π(ab)cosec(πs)\boxed{\displaystyle F(a)=\pi (a-b)\mathrm{cosec}\left(\dfrac{\pi}{s}\right)}

Proof of:

0+11+xsdx=πscosec(πs)\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)

Perform the change of variable

y=xsy=x^s

0+11+xsdx=1s0+x1s11+xdx\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{1}{s} \int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx

0+x1s11+xdx=β(1s,11s)=Γ(1s)Γ(11s)Γ(1)=πsin(πs)=πcosec(πs) \displaystyle \begin{aligned}\int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx&=\beta\left(\dfrac{1}{s},1-\dfrac{1}{s}\right)\\ &=\dfrac{\Gamma\left(\dfrac{1}{s}\right)\Gamma\left(1-\dfrac{1}{s}\right)}{\Gamma(1)}\\ &= \dfrac{\pi}{\sin\left(\dfrac{\pi}{s}\right)}\\ &=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right) \end{aligned}

β\beta being the Beta Euler function. Third line is the use of Euler's reflection formula.

FDP DPF - 4 years, 5 months ago

Problem 25 :

Prove That

1π0πxarctan(psinx1+pcosx) dx=Li2(p) p<1 \dfrac{1}{\pi} \int_{0}^{\pi} x \arctan\left( \dfrac{p \sin x}{1+p \cos x} \right) \ \mathrm{d}x = \operatorname{Li}_{2}(p) \quad \forall \ |p| <1

Notation : Li2(z)\operatorname{Li}_{2}(z) denotes the Dilogarithm Function.

This problem has been solved by Mark Hennings.

Ishan Singh - 4 years, 5 months ago

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If we define g(p)  =  0πln(1+2pcosx+p2)dxp<1 g(p) \; = \; \int_0^\pi \ln(1 + 2p \cos x + p^2)\,dx \hspace{1cm} |p| < 1 then g(0)=0g(0) = 0 and, using the complex subsitution z=eixz = e^{ix}, g(p)=0π2cosx+2p1+2pcosx+p2dx  =  12ππ2cosx+2p1+2pcosx+p2dx=12z=1z+z1+2p1+p(z+z1)+p2dziz  =  12iz=1z2+2pz+1z(pz+1)(z+p)dz=π(Resz=0+Resz=p)z2+2pz+1z(pz+1)(z+p)=π(1p1p2p(1p2))  =  0\begin{aligned} g'(p) & = \int_0^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \; = \; \frac12\int_{-\pi}^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{2}\int_{|z|=1} \frac{z + z^{-1} + 2p}{1 + p(z + z^{-1}) + p^2} \frac{dz}{iz} \; = \; \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 2pz + 1}{z(pz+1)(z+p)}\,dz \\ & = \pi\left(\mathrm{Res}_{z=0} + \mathrm{Res}_{z=-p}\right)\frac{z^2 + 2pz + 1}{z(pz+1)(z+p)} \\ & = \pi\left(\frac{1}{p} - \frac{1-p^2}{p(1-p^2)}\right) \; = \; 0 \end{aligned} so that g(p)=0g(p) = 0 for all p<1|p| < 1. If we now define f(p)  =  1π0πxtan1(psinx1+pcosx)dxp<1 f(p) \; = \; \frac{1}{\pi}\int_0^\pi x \tan^{-1}\left(\frac{p \sin x}{1 + p \cos x}\right)\,dx \hspace{1cm} |p| < 1 then f(0)=0f(0) = 0 and f(p)=1π0πx1+(psinx1+pcosx)2×sinx(1+pcosx)2dx=1π0πxsinx1+2pcosx+p2dx=1π[x2pln(1+2pcosx+p2)]0π+12πp0πln(1+2pcosx+p2)dx=ln(1p)p+g(p)  =  ln(1p)p\begin{aligned} f'(p) & = \frac{1}{\pi}\int_0^\pi \frac{x}{1 + \left(\frac{p \sin x}{1 + p\cos x}\right)^2} \times \frac{\sin x}{(1 + p\cos x)^2}\,dx \\ & = \frac{1}{\pi}\int_0^\pi \frac{x \sin x}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{\pi}\Big[-\frac{x}{2p}\ln(1 + 2p\cos x + p^2)\Big]_0^\pi + \frac{1}{2\pi p}\int_0^\pi \ln(1 + 2p\cos x + p^2)\,dx \\ & = -\frac{\ln(1-p)}{p} + g(p) \; = \; -\frac{\ln(1-p)}{p} \end{aligned} and hence it follows that f(p)=Li2(p)f(p) \,=\, \mathrm{Li}_2(p).

Mark Hennings - 4 years, 5 months ago

Aliter,

Let f(p)=1π0πxarctan(psinx1+pcosx) dxf(p) = \dfrac{1}{\pi}\int_{0}^{\pi} x \arctan\left( \dfrac{p\sin x}{1+ p\cos x} \right) \ \mathrm{d}x

Firstly, we have f(0)=0f(0) = 0

Note that,

k=1(p)k1sin(kx)=1p(k=0(peix)k) \displaystyle \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx) = \dfrac{1}{p}\Im \left( \sum_{k=0}^{\infty} (-p e^{ix})^{k} \right)

=1p(11+peix)\displaystyle = \dfrac{1}{p} \Im \left( \dfrac{1}{1+p e^{ix}} \right)

=sinxp2+2pcosx+1\displaystyle = \dfrac{\sin x}{p^2 +2p \cos x +1}

where (z)\Im(z) denotes the imaginary part of zz.

So we have,

f(p)=1π0πxsinxp2+2pcosx+1 dx \displaystyle f'(p) = \dfrac{1}{\pi} \int_{0}^{\pi} \dfrac{x \sin x}{p^2 +2p \cos x +1} \ \mathrm{d}x

=1π0πx(k=1(p)k1sin(kx)) dx \displaystyle = \dfrac{1}{\pi} \int_{0}^{\pi} x \left(\sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)\right) \ \mathrm{d}x

=1πk=1(p)k10πxsin(kx) dx \displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \int_{0}^{\pi}x \sin (kx) \ \mathrm{d}x

=1πk=1(p)k1(sin(kπ)kπcos(kπ)k2) \displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \left( \dfrac{\sin(k \pi) - k \pi \cos(k \pi)}{k^2} \right)

Since sin(kπ)=0\sin (k \pi) = 0 and cos(kπ)=(1)k  kZ\cos (k \pi) = (-1)^k \ \forall \ k \in \mathbb{Z}, we get,

f(p)=k=1pk1k=ln(1p)p \displaystyle f'(p) = \sum_{k=1}^{\infty} \dfrac{p^{k-1}}{k} = -\dfrac{\ln(1-p)}{p}

f(p)=Li2(p)  \displaystyle \therefore f(p) = \operatorname{Li}_{2}(p) \ \square

Ishan Singh - 4 years, 5 months ago

Problem 27:

Show that

062162+1lnx(x1)x22(15+83)x+1dx=23(23)G\displaystyle\int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}\, dx=\dfrac{2}{3}(2-\sqrt{3})G

This problem has been solved by Mark Hennings.

FDP DPF - 4 years, 5 months ago

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Put α=112π\alpha = \tfrac{1}{12}\pi. With the substitution x=1cosθ1+cosθx = \frac{1-\cos\theta}{1+\cos\theta}, we obtain after much manipulation that I  =  062162+1lnxdx(x1)x22(15+83)x+1)  =  12sinα0αln(1cosθ1+cosθ)sinθdθcosθcos2θcos2α I \; = \; \int_0^{\frac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \frac{\ln x\,dx}{(x-1)\sqrt{x^2 - 2(15+8\sqrt{3})x + 1)}} \; = \; -\tfrac12\sin\alpha \int_0^\alpha \frac{\ln\left(\frac{1-\cos\theta}{1+\cos\theta}\right) \sin\theta\,d\theta}{\cos\theta\sqrt{\cos^2\theta - \cos^2\alpha}} Putting u=cosθu = \cos\theta now gives I  =  12sinαcosα1ln(1u1+u)duuu2cos2α I \; = \; -\tfrac12\sin\alpha \int_{\cos\alpha}^1 \frac{\ln\left(\frac{1-u}{1+u}\right)\,du}{u\sqrt{u^2 - \cos^2\alpha}} Now putting u=cosαsecϕu = \cos\alpha \sec\phi yields I=12tanα0αln(cosϕcosαcosϕ+cosα)dϕ  =  12tanα0αln(tan(α+ϕ2)tan(αϕ2))dϕ=12tanα0α{ln(tan(α+ϕ2))+ln(tan(αϕ2))}dϕ=tanα{12ααln(tanϕ)dϕ12α0ln(tanϕ)dϕ}=tanα0αln(tanϕ)dϕ  =  tanα0112πln(tanϕ)dϕ=23tanαG  =  23(23)G\begin{aligned} I & = -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\frac{\cos\phi - \cos\alpha}{\cos\phi + \cos\alpha}\right)\,d\phi \; = \; -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big) \tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\,d\phi \\ & = -\tfrac12\tan\alpha \int_0^\alpha \left\{ \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big)\right) + \ln\left(\tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\right\}\,d\phi \\ & = -\tan\alpha \left\{ \int_{\frac12\alpha}^\alpha \ln(\tan \phi)\,d\phi - \int_{\frac12\alpha}^0 \ln(\tan\phi)\,d\phi\right\} \\ & = -\tan\alpha \int_0^\alpha \ln(\tan\phi)\,d\phi \; = \; -\tan\alpha \int_0^{\frac{1}{12}\pi}\ln(\tan\phi)\,d\phi \\ & = \tfrac23\tan\alpha\, G \; = \; \tfrac23(2 - \sqrt{3})G \end{aligned} using a standard integral representation of GG at the very last stage.

Mark Hennings - 4 years, 5 months ago

Problem 29:

Show that 0etxx(x2+1)dx=π[2C(2tπ)sint2S(2tπ)costsin(tπ/4)]\displaystyle \int\limits_{0}^{\infty} \frac{e^{-tx}}{\sqrt x (x^2+1)}dx=\pi \left[ \sqrt2 C\left( \sqrt{\frac{2t}{\pi}}\right)\sin t - \sqrt2S\left( \sqrt{\frac{2t}{\pi}} \right) \cos t - \sin(t-\pi/4) \right]

where CC and SS are Fresnel cosine integral and Fresnel sine integral, respectively.

This problem has been solved by Mark Hennings.

Aman Rajput - 4 years, 5 months ago

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In this question, we are using the Fresnel integrals C(x)  =  0xcos(12πt2)dtS(x)  =  0xsin(12πt2)dt C(x) \; = \; \int_0^x \cos\big(\tfrac12\pi t^2\big)\,dt \hspace{2cm} S(x) \; = \; \int_0^x \sin\big(\tfrac12\pi t^2\big)\,dt If we define A(t)  =  cost0etxxx2+1dxsint0etxx(x2+1)dx A(t) \; = \; \cos t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx for t>0t > 0 then A(t)=[t]lsint0etxxx2+1dxcost0etxxxx2+1dxcost0etxx(x2+1)dx+sint0etxxx2+1dx=cost0etxxdx  =  πtcost\begin{aligned} A'(t) & = \begin{array}{c}[t]{l}\displaystyle-\sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \cos t \int_0^\infty \frac{e^{-tx}x\sqrt{x}}{x^2+1}\,dx \\\displaystyle - \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx + \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx\end{array} \\ & = -\cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; = \; -\sqrt{\tfrac{\pi}{t}}\cos t \end{aligned} for any t>0t > 0, so that ddt[A(t)+π2C(2tπ)]  =  0 \frac{d}{dt}\Big[ A(t) + \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\Big] \; = \; 0 Since A(t)0A(t) \,\to\, 0 as tt \to \infty, we deduce that A(t)+π2C(2tπ)  =  12πt>0 A(t) + \pi\sqrt{2}C\Big(\sqrt{\frac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0 Similarly, if we define B(t)  =  sint0etxxx2+1dx+cost0etxx(x2+1)dx B(t) \; = \; \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx + \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx for t>0t > 0, then we can show that B(t)  =  sint0etxxdx  =  πtsint B'(t) \; =\; -\sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; =\; - \sqrt{\tfrac{\pi}{t}}\sin t and hence B(t)+π2S(2tπ)  =  12πt>0 B(t) + \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0 Thus we deduce that 0etxx(x2+1)dx=B(t)costA(t)sint=cost[12ππ2S(2tπ)]sint[12ππ2C(2tπ)]=π2C(2tπ)π2S(2tπ)πsin(t14π)\begin{aligned} \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx & = B(t)\cos t - A(t) \sin t \\ & = \cos t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] - \sin t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] \\ & = \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sin\big(t - \tfrac14\pi\big) \end{aligned} as required.

Mark Hennings - 4 years, 5 months ago

Problem 30:

Show that 01ln(1+14x2+12x)dxx  =  120π2 \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\,\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2

This problem has been solved by Fdp Dpf.

Mark Hennings - 4 years, 5 months ago

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I=01ln(1+x24+x2)xdx\displaystyle I=\int_0^1 \dfrac{\ln(\sqrt{1+\dfrac{x^2}{4}}+\dfrac{x}{2})}{x}dx

Perform the change of variable y=x2y=\dfrac{x}{2},

I=012ln(1+x2+x)xdx\displaystyle I=\int_0^{\tfrac{1}{2}} \dfrac{\ln\left(\sqrt{1+x^2}+x\right)}{x}dx

Perform the change of variable y=arsinh(x)y=\mathrm{arsinh}(x),

I=0arsinh(12)xcosh(x)sinh(x)dx=0arsinh(12)x(1+e2x)1e2xdx=0arsinh(12)(x(1+e2x)n=0+e2nx)dx=(arsinh(12))22+20arsinh(12)(xn=1+e2nx)dx=(arsinh(12))22+2n=1+(0arsinh(12)xe2nx)dx=(arsinh(12))22+12n=1+(1n2(1+2arsinh(12)n)e2arsinh(12)nn2)=(arsinh(12))22+12ζ(2)12Li2(e2asinh(12))+arsinh(12)ln(1e2arsinh(12))\begin{aligned}\displaystyle I&=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\cosh(x)}{\sinh(x)}dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\left(1+\mathrm{e}^{-2x}\right)}{1-\mathrm{e}^{-2x}} dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\left(1+\mathrm{e}^{-2x}\right)\sum_{n=0}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\sum_{n=1}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\sum_{n=1}^{+\infty} \left(\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} x\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\sum_{n=1}^{+\infty}\left(\dfrac{1}{n^2}-\dfrac{\left(1+2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n\right)\mathrm{e}^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n}}{n^2}\right)\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\zeta(2)-\dfrac{1}{2}\mathrm{Li}_2\left(e^{-2\mathrm{asinh}\left(\tfrac{1}{2}\right)}\right)+\mathrm{arsinh}\left(\tfrac{1}{2}\right)\ln\left(1-e^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)}\right) \end{aligned}

Since, arsinh(12)=ln(12+52)\mathrm{arsinh}\left(\dfrac{1}{2}\right)=\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right) then,

I=π21212Li2(3252)12(ln(12+52))2I=\dfrac{\pi^2}{12}-\dfrac{1}{2}\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)-\dfrac{1}{2}\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2

Since , Li2(3252)=π215(ln(12+52))2\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=\dfrac{\pi^2}{15}-\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2 then,

I=π212π230=π220I=\dfrac{\pi^2}{12}-\dfrac{\pi^2}{30}=\boxed{\dfrac{\pi^2}{20}}

FDP DPF - 4 years, 5 months ago

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Nice. A bit of integration by parts makes the earlier stage a little clearer, perhaps...

Note that ddxln(1+14x2+12x)  =  12x121+14x2+121+14x2+12x  =  14+x2 \frac{d}{dx}\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right) \; = \; \frac{\frac{\frac12x}{\frac12\sqrt{1+\frac14x^2}} + \frac12}{\sqrt{1 + \frac14x^2} + \frac12x} \; = \; \frac{1}{\sqrt{4+x^2}} and so, integrating by parts, 01ln(1+14x2+12x)dxx=[(lnx)ln(1+14x2+12x)]0101lnx4+x2dx=01lnx4+x2dx\begin{aligned} \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} & = \Big[(\ln x)\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\Big]_0^1 - \int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \\ & = -\int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \end{aligned} Since sinh112=ln(12(5+1))\sinh^{-1}\tfrac12 = \ln\big(\tfrac12(\sqrt{5}+1)\big), the substitution x=2sinhux = 2\sinh u yields 01lnx4+x2dx=0sinh112ln(2sinhu)du  =  0sinh112[u+ln(1e2u)]du=12ln2(5+12)+0ln(5+12)ln(1e2u)e2ue2udu=12ln2(5+12)121352ln(1v)vdv\begin{aligned} \int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx & = \int_0^{\sinh^{-1}\frac12} \ln\big(2\sinh u\big)\,du \; = \; \int_0^{\sinh^{-1}\frac12}\Big[u + \ln\big(1 - e^{-2u}\big)\Big]\,du \\ & = \tfrac12 \ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \int_0^{\ln\big(\frac{\sqrt{5}+1}{2}\big)} \frac{\ln\big(1 - e^{-2u}\big)}{e^{-2u}}\,e^{-2u}\,du \\ & = \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) - \frac12\int_1^{\frac{3-\sqrt{5}}{2}} \frac{\ln(1-v)}{v}\,dv \end{aligned} using the substitution v=e2uv = e^{-2u}. Thus 01lnx4+x2dx  =  12ln2(5+12)+12Li2(352)12Li2(1)  =  130π2112π2  =  120π2 \int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx \; = \; \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \tfrac12\mathrm{Li}_2\left(\tfrac{3 - \sqrt{5}}{2}\right) - \tfrac12\mathrm{Li}_2(1) \; = \; \tfrac{1}{30}\pi^2 - \tfrac{1}{12}\pi^2 \; = \; -\tfrac{1}{20}\pi^2 and so 01ln(1+14x2+12x)dxx  =  120π2 \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2

Mark Hennings - 4 years, 5 months ago

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@Mark Hennings You're right, it's easiest this way. By the way, ln(1+14x2+12x)dx=xln(1+14x2+12x)x2+4\displaystyle \int \ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)dx=x\ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)-\sqrt{x^2+4}

FDP DPF - 4 years, 5 months ago

Problem 31:

Show that

01x1(x+1)lnxdx=ln(π2)\displaystyle \int_0^1 \dfrac{x-1}{(x+1)\ln x}dx=\ln\left(\dfrac{\pi}{2}\right)

This problem has been solved by Ishan Singh.

FDP DPF - 4 years, 5 months ago

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Let

f(p)=01xp1(x+1)lnx dx;p>0f(p) = \int_{0}^{1} \dfrac{x^p - 1}{(x+1) \ln x} \ \mathrm{d}x \quad ; \quad p > 0

    f(p)=01xpx+1dx \displaystyle \implies f'(p) = \int_{0}^{1} \dfrac{x^p}{x+1} \mathrm{d}x

=01xp(x1)x21dx\displaystyle = \int_{0}^{1} \dfrac{x^p(x-1)}{x^2 -1} \mathrm{d}x

Substitute x2xx^2 \mapsto x and simplify to get,

f(p)=1201(xp21x1xp121x1)dx \displaystyle f'(p) = \dfrac{1}{2} \int_{0}^{1} \left( \dfrac{x^{\frac{p}{2}} - 1}{x - 1} - \dfrac{x^{\frac{p-1}{2}} - 1}{x - 1} \right) \mathrm{d}x

=12(ψ(p2+1)ψ(p12+1))\displaystyle = \dfrac{1}{2} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right)

Note that f(0)=0f(0) = 0

    f(1)=01f(p) dp\displaystyle \implies f(1) = \int_{0}^{1} f'(p) \ \mathrm{d}p

=1201(ψ(p2+1)ψ(p12+1))dp\displaystyle = \dfrac{1}{2} \int_{0}^{1} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right) \mathrm{d}p

=log(π2)\displaystyle = \log \left(\dfrac{\pi}{2}\right)

where the last equality follows since ψ(z)=ddzlog(Γ(z)) \psi (z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \log(\Gamma(z))

Ishan Singh - 4 years, 5 months ago

Problem 37:

Evaluate 012πln(cosx+sinx)dx\int_0^{\frac12\pi}\, \ln(\cos x + \sin x)\,dx

This problem has been solved by Fdp Dpf.

Mark Hennings - 4 years, 5 months ago

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Solution to problem 37.

Since,

sin(x+π4)=cos(π4)sinx+sin(π4)cosx=22(sinx+cosx)\begin{aligned}\sin\left(x+\dfrac{\pi}{4}\right)&=\cos\left(\dfrac{\pi}{4}\right)\sin x+\sin \left(\dfrac{\pi}{4}\right)\cos x\\ &=\dfrac{\sqrt{2}}{2}(\sin x+\cos x)\end{aligned}

then,

0π2ln(cosx+sinx)dx=0π2ln(2sin(x+π4))dx=0π4ln(2sin(x+π4))dx+π4π2ln(2sin(x+π4))dx=0π4ln(2sin((π4x)+π4))dx+0π4ln(2sin((x+π4)+π4))dx=20π4ln(2cosx)dx=πln24+20π4ln(cosx)dx=πln240π4ln(1(cosx)2)dx=πln240π4ln(1+(tanx)2)dx\begin{aligned} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\int_0^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(\dfrac{\pi}{4}-x\right)+\dfrac{\pi}{4}\right)\right)dx+\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(x+\dfrac{\pi}{4}\right)+\dfrac{\pi}{4}\right)\right)dx\\ &=2\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}+2\int_0^{\tfrac{\pi}{4}} \ln\left(\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(\dfrac{1}{(\cos x)^2}\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(1+(\tan x)^2\right)dx\\ \end{aligned}

In the latter integral perform the change of variable y=tanxy=\tan x, therefore, 0π2ln(cosx+sinx)dx=πln2401ln(1+x2)1+x2dx\begin{aligned} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx \end{aligned}

0+ln(1+x2)1+x2dx=01ln(1+x2)1+x2dx+0+ln(1+x2)1+x2dx\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx

In the latter integral perform the change of variable y=1xy=\dfrac{1}{x}, therefore,

0+ln(1+x2)1+x2dx=01ln(1+x2)1+x2dx+01ln(1+1x2)1+x2dx=201ln(1+x2)1+x2dx201lnx1+x2dx=201ln(1+x2)1+x2dx+2G\begin{aligned}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx&=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^1 \dfrac{\ln\left(1+\dfrac{1}{x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx-2\int_0^1 \dfrac{\ln x}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+2\text{G}\\ \end{aligned}

G\text{G} being the Catalan constant. But, 0+ln(1+x2)1+x2dx=πln2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\pi\ln 2

(see my answer to problem 36)

Therefore, 01ln(1+x2)1+x2dx=πln22G\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{2}-\text{G}

Therefore,

0π2ln(cosx+sinx)dx=πln24(πln22G)=Gπln24\begin{aligned} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\left(\dfrac{\pi\ln 2}{2}-\text{G}\right)\\ &=\boxed{\text{G}-\dfrac{\pi\ln 2}{4}} \end{aligned}

FDP DPF - 4 years, 5 months ago

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Aliter,

The integral 0π4log(cosx) dx \displaystyle \int_{0}^{\frac{\pi}{4}} \log(\cos x) \ \mathrm{d}x can also be done using the identity log(cosx)=log2k=1cos(2kx)k \log(\cos x) = -\log 2 - \sum_{k=1}^{\infty} \dfrac{\cos (2kx)}{k}.

Ishan Singh - 4 years, 5 months ago

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@Ishan Singh I know but i don't like to use this identity when i can avoid to use it. I like solutions using elementary tools.

FDP DPF - 4 years, 5 months ago

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@Fdp Dpf I like solutions using elementary methods too. I consider the above identity elementary, it can be proved by writing cos(2kx)\cos (2kx) as e2ikx+e2ikx2\dfrac{e^{2ikx} + e^{-2ikx}}{2} and then using the Taylor Series of logarithm. Btw, Happy New Year.

Ishan Singh - 4 years, 5 months ago

Problem 28:

Show that,

01arctan(x(1x)2x)xdx=G3\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}\, dx=\frac{G}{3}

GG being the Catalan constant.

This problem has been solved by Mark Hennings.

FDP DPF - 4 years, 5 months ago

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Your derivation of this result on MSE is pretty compact, so I shall expand it. Suppose that a>1a > 1, and let b=1+a2a1b = \frac{1+a^2}{a-1}. The substitution y=axx+by \,=\, \frac{ax}{x+b} gives 01tan1(axx+b)dxx  =  0a1a+1tan1yaybyab(ay)2dy  =  0a1a+1ay(ay)tan1ydy \int_0^1 \tan^{-1}\left(\frac{ax}{x+b}\right)\,\frac{dx}{x} \; = \; \int_0^{\frac{a-1}{a+1}} \tan^{-1}y\, \frac{a-y}{by} \, \frac{ab}{(a-y)^2}\,dy \; = \; \int_0^{\frac{a-1}{a+1}} \frac{a}{y(a-y)} \tan^{-1}y\,dy while the substitution z=xbaxz = \frac{x}{b - ax} gives 01tan1(xaxb)dxx  =  0a1a+1tan1zaz+abzb(az+1)2dz  =  0a1a+11z(az+1)tan1zdz \int_0^1 \tan^{-1}\left(\frac{x}{ax - b}\right)\,\frac{dx}{x} \; = \; -\int_0^{\frac{a-1}{a+1}} \tan^{-1}z \, \frac{az+a}{bz} \, \frac{b}{(az+1)^2}\,dz \; =\; -\int_0^{\frac{a-1}{a+1}} \frac{1}{z(az+1)} \tan^{-1}z\,dz and hence 01{tan1(axax+b)+tan1(xaxb)}dxx=0a1a+1(ay(ay)1y(ay+1))tan1ydy=0a1a+1a2+1(ay)(ay+1)tan1ydy=)a1a+1(1ay+a1+ay)tan1ydy=[ln(1+ayay)tan1y]0a1a+1+0a1a+1ln(ay1+ay)1+y2dy=0a1a+1ln(ay1+ay)1+y2dy  =  1alnx1+x2dx\begin{aligned} \int_0^1 \left\{\tan^{-1}\left(\frac{ax}{ax+b}\right) + \tan^{-1}\left(\frac{x}{ax-b}\right)\right\}\,\frac{dx}{x} & = \int_0^{\frac{a-1}{a+1}} \left(\frac{a}{y(a-y)} - \frac{1}{y(ay+1)}\right) \tan^{-1}y\,dy \\ & = \int_0^{\frac{a-1}{a+1}} \frac{a^2+1}{(a-y)(ay+1)}\,\tan^{-1}y\,dy \\ & = \int_)^{\frac{a-1}{a+1}} \left(\frac{1}{a-y} + \frac{a}{1 + ay}\right)\tan^{-1}y\,dy \\ & = \left[ \ln\left(\frac{1 + ay}{a-y}\right) \tan^{-1}y\right]_0^{\frac{a-1}{a+1}} + \int_0^{\frac{a-1}{a+1}} \frac{\ln\left(\frac{a-y}{1+ay}\right)}{1+y^2}\,dy \\ & = \int_0^{\frac{a-1}{a+1}} \frac{\ln\left(\frac{a-y}{1+ay}\right)}{1+y^2}\,dy \; = \; \int_1^a \frac{\ln x}{1+x^2}\,dx \end{aligned} Now tan[tan1(axx+b)+tan1(xaxb)]=axx+b+xaxb1axx+bxaxb  =  ax(axb)+x(x+b)(x+b)(axb)ax2=x(1x)1+a2(a1)2x\begin{aligned} \tan\left[\tan^{-1}\left(\frac{ax}{x+b}\right) + \tan^{-1}\left(\frac{x}{ax-b}\right)\right] & = \frac{\frac{ax}{x+b} + \frac{x}{ax-b}}{1 - \frac{ax}{x+b}\frac{x}{ax-b}} \; = \; \frac{ax(ax-b) + x(x+b)}{(x+b)(ax-b) - ax^2} \\ & = \frac{x(1-x)}{\frac{1+a^2}{(a-1)^2} - x} \end{aligned} and hence it follows that 01tan1(x(1x)1+a2(a1)2x)dxx  =  1alnx1+x2dx \int_0^1 \tan^{-1}\left(\frac{x(1-x)}{\frac{1+a^2}{(a-1)^2} - x}\right)\,\frac{dx}{x} \; = \; \int_1^a \frac{\ln x}{1 + x^2}\,dx Putting a=2+3a = 2 + \sqrt{3} yields 01tan1(x(1x)2x)dxx=12+3lnx1+x2dx  =  14π512πln(tanθ)dθ=112π14πln(tanϕ)dϕ  =  ((G)(23G))  =  13G\begin{aligned} \int_0^1 \tan^{-1}\left(\frac{x(1-x)}{2-x}\right)\,\frac{dx}{x} & = \int_1^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx \; = \; \int_{\frac14\pi}^{\frac{5}{12}\pi} \ln(\tan\theta)\,d\theta \\ & = -\int_{\frac{1}{12}\pi}^{\frac14\pi} \ln(\tan\phi)\,d\phi \; = \; -\big((-G) - (-\tfrac23G)\big) \; = \; \tfrac13G \end{aligned} using the final substitution ϕ=12πθ\phi = \tfrac12\pi - \theta.

I won't be posting the next problem until after Christmas!

Mark Hennings - 4 years, 5 months ago

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Nice. I was secretly expecting a different proof. Someone, in 2011, showed me this integral. I have tried to find out the very source for it. Since then, i enjoy to learn and sharing knowledge about computing integrals. Happy chrismas !

FDP DPF - 4 years, 5 months ago

Problem 36:

Show that,

0+xln(1+2x+x2)1+x4dx=πln22 \int_0^{+\infty}\dfrac{x\ln(1+\sqrt{2}x+x^2)}{1+x^4}dx=\dfrac{\pi\ln 2}{2}

This problem has been solved by Mark Hennings.

FDP DPF - 4 years, 5 months ago

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The substitutions t=1st = 1-s and s=sinθs = \sin\theta give 01(1t(2t)1)dt1t  =  01(11s21)dss  =  012πtan12θdθ  =  ln2 \int_0^1 \left(\frac{1}{\sqrt{t(2-t)}}-1\right)\,\frac{dt}{1-t} \; = \; \int_0^1 \left(\frac{1}{\sqrt{1-s^2}} - 1\right)\,\frac{ds}{s} \; = \; \int_0^{\frac12\pi}\tan\tfrac12\theta\,d\theta \; = \; \ln2 The substitution x2=t2tx^2 = \tfrac{t}{2-t} and one of the standard integral representations of the Catalan constant GG gives 011t(2t)(14πtan1t2t)dt1t  =  201(14πtan1x)dx1x2  =  G \int_0^1 \frac{1}{\sqrt{t(2-t)}} \left(\tfrac14\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right)\,\frac{dt}{1-t} \; = \; 2\int_0^1 \left(\tfrac14\pi - \tan^{-1}x\right)\,\frac{dx}{1-x^2} \; = \; G These results together give us that 0ln(x2+2x+1)x2+2x+1dx=0dxx2+2x+101x2+2xtx2+t2x+1dt=010x2+2x(tx2+t2x+1)(x2+2x+1)dxdt=01dt1t0(1tx2+t2x+11x2+2x+1)dx=01dt1t{2t(2t)(12πtan1t2t)π22}=01dt1t{2t(2t)(14πtan1t2t)+π22(1t(2t)1)}\begin{aligned} \int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\,dx & = \int_0^\infty \frac{dx}{x^2 + \sqrt{2}x + 1}\int_0^1 \frac{x^2 + \sqrt{2}x}{tx^2 + t\sqrt{2}x + 1}\,dt \\ & = \int_0^1 \int_0^\infty \frac{x^2 + \sqrt{2}x}{(tx^2 + t\sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}\,dx\,dt \\ & = \int_0^1 \frac{dt}{1-t}\int_0^\infty\left(\frac{1}{tx^2 + t\sqrt{2}x + 1} - \frac{1}{x^2 + \sqrt{2}x + 1}\right)\,dx \\ & = \int_0^1 \frac{dt}{1-t} \left\{ \sqrt{\tfrac{2}{t(2-t)}}\left(\tfrac12\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right) - \frac{\pi}{2\sqrt{2}}\right\} \\ & = \int_0^1 \frac{dt}{1-t}\left\{ \sqrt{\tfrac{2}{t(2-t)}}\left(\tfrac14\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right) + \tfrac{\pi}{2\sqrt{2}}\left(\frac{1}{\sqrt{t(2-t)}} - 1\right)\,\right\} \end{aligned} and so 0ln(x2+2x+1)x2+2x+1dx  =  =2G+π22ln2 \int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\,dx \; = \; = \sqrt{2}G + \tfrac{\pi}{2\sqrt{2}}\ln2 Standard substitutions show that f(a)  =  0x2+1(x4+1)adx  =  14[B(34,a34)+B(14,a14)] f(a) \; = \; \int_0^\infty \frac{x^2+1}{(x^4 + 1)^a}\,dx \; = \; \tfrac14\big[B(\tfrac34,a-\tfrac34)+B(\tfrac14,a-\tfrac14)\big] for a>34a > \tfrac34, and hence f(a)  =  Γ(34)Γ(a34)ψ(a34)+Γ(14)Γ(a14)ψ(a14)4Γ(a)Γ(34)Γ(a34)+Γ(14)Γ(a14)4Γ(a)ψ(a) f'(a) \; = \; \frac{\Gamma(\frac34)\Gamma(a-\frac34)\psi(a-\frac34) + \Gamma(\tfrac14)\Gamma(a-\frac14)\psi(a-\frac14)}{4\Gamma(a)} - \frac{\Gamma(\frac34)\Gamma(a-\frac34) + \Gamma(\frac14)\Gamma(a-\frac14)}{4\Gamma(a)}\psi(a) and hence 0(x2+1)ln(x4+1)x4+1dx  =  f(1)  =  π22[ψ(14)+ψ(34)2ψ(1)]  =  3π2ln2 \int_0^\infty \frac{(x^2+1)\ln(x^4+1)}{x^4+1}\,dx \; = \; -f'(1) \; = \; -\tfrac{\pi}{2\sqrt{2}}\big[\psi(\tfrac14) + \psi(\tfrac34) - 2\psi(1)\big] \; = \; \tfrac{3\pi}{\sqrt{2}}\ln2 Next, the substitution x=tan12θx = \tan\tfrac12\theta and another standard integral representation of the Catalan constant gives 01+x21+x4ln(x2+2x+1x22x+1)dx=0πln(1+12sinθ112sinθ)dθ1+cos2θ=2012πln(1+12sinθ112sinθ)dθ1+cos2θ  =  22G\begin{aligned} \int_0^\infty &\frac{1+x^2}{1+x^4}\ln\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right)\,dx \\ & = \int_0^\pi \ln\left(\frac{1 + \frac{1}{\sqrt{2}}\sin\theta}{1 - \frac{1}{\sqrt{2}}\sin\theta}\right) \frac{d\theta}{1 + \cos^2\theta} \\ & = 2\int_0^{\frac12\pi} \ln\left(\frac{1 + \frac{1}{\sqrt{2}}\sin\theta}{1 - \frac{1}{\sqrt{2}}\sin\theta}\right) \frac{d\theta}{1 + \cos^2\theta} \; = \; 2\sqrt{2}G \end{aligned} Thus 0(ln(x2+2x+1)x2+2x+1+ln(x2+2x+1)x22x+1)dx=20x2+1x4+1ln(x2+2x+1)dx=0x2+1x4+1[ln(x2+2x+1x22x+1)+ln(x4+1)]dx=22G+3π2ln2\begin{aligned} \int_0^\infty &\left(\frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1} + \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 - \sqrt{2}x + 1}\right)\,dx \\ & = 2\int_0^\infty \frac{x^2+1}{x^4+1} \ln(x^2+ \sqrt{2}x+1)\,dx \\ & = \int_0^\infty \frac{x^2+1}{x^4+1}\left[\ln\left(\frac{x^2 +\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right) + \ln(x^4+1)\right]\,dx \\ & = 2\sqrt{2}G + \tfrac{3\pi}{\sqrt{2}}\ln2 \end{aligned} and hence 0ln(x2+2x+1)x22x+1dx  =  (22G+3π2ln2)(2G+π22ln2)  =  2G+5π22ln2 \int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 - \sqrt{2}x + 1}\,dx \; = \; \big( 2\sqrt{2}G + \tfrac{3\pi}{\sqrt{2}}\ln2 \big) - \big( \sqrt{2}G + \tfrac{\pi}{2\sqrt{2}}\ln2\big) \; = \; \sqrt{2}G + \tfrac{5\pi}{2\sqrt{2}}\ln2 Finally (!) we deduce that 0xx4+1ln(x2+2x+1)dx=1220(ln(x2+2x+1)x22x+1ln(x2+2x+1)x2+2x+1)=12πln2\begin{aligned} \int_0^\infty \frac{x}{x^4+1}\ln(x^2 + \sqrt{2}x + 1)\,dx & = \frac{1}{2\sqrt{2}}\int_0^\infty\left(\frac{\ln(x^2 + \sqrt{2}x+1)}{x^2 - \sqrt{2}x + 1} - \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\right) \\ & = \tfrac12\pi \ln2 \end{aligned}

Mark Hennings - 4 years, 5 months ago

Alternative solution to problem 36

Let,

I=0+x(ln(1+2x+x2)1+x4dxI=\int_0^{+\infty} \dfrac{x(\ln(1+\sqrt{2}x+x^2)}{1+x^4}dx

J=0+x(ln(12x+x2)1+x4dxJ=\int_0^{+\infty} \dfrac{x(\ln(1-\sqrt{2}x+x^2)}{1+x^4}dx

I+J=0+xln(1+x4)1+x4dxI+J=\int_0^{+\infty} \dfrac{x\ln(1+x^4)}{1+x^4}dx

Perform the change of variable y=x2y=x^2,

I+J=120+ln(1+x2)1+x2dxI+J=\dfrac{1}{2}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx

Perform the change of variable y=arctanxy=\arctan x,

I+J=0π2ln(cosx)dxI+J=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx

Perform the change of variable y=π2xy=\dfrac{\pi}{2}-x,

0π2ln(cosx)dx=0π2ln(sinx)dx\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx=\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx

Therefore,

2(I+J)=0π2ln(cosx)dx0π2ln(sinx)dx=0π2ln(cos(x)sin(x))dx=0π2ln(sin(2x)2)dx=0π2ln(sin(2x))dx+πln22\begin{aligned} 2(I+J)&=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx-\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\cos(x)\sin(x) )dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln\left(\dfrac{\sin(2x)}{2} \right)dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln\left(\sin(2x) \right)dx+\dfrac{\pi\ln 2}{2}\\ \end{aligned}

In the latter integral perform the change of variable y=2xy=2x,

2(I+J)=120πln(sinx)dx+πln22=120π2ln(sinx)dx12π2πln(sinx)dx+πln22\begin{aligned} 2(I+J)&=-\dfrac{1}{2}\int_0^{\pi} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx-\dfrac{1}{2}\int_{\tfrac{\pi}{2}}^{\pi} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ \end{aligned}

In the latter integral perform the change of variable y=πxy=\pi-x, therefore,

2(I+J)=120π2ln(sinx)dx120π2ln(sinx)dx+πln22=0π2ln(sinx)dx+πln22=0π2ln(cosx)dx+πln22=I+J+πln22\begin{aligned} 2(I+J)&=-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx+\dfrac{\pi\ln 2}{2}\\ &=I+J+\dfrac{\pi\ln 2}{2}\\ \end{aligned}

Therefore,

I+J=πln22\boxed{I+J=\dfrac{\pi\ln 2}{2}}

Define on [0;2]\Big[0;\sqrt{2}\Big],

F(a)=0+x(ln(1+ax+x2)ln(1ax+x2))1+x4dx\displaystyle F(a)= \int_0^{+\infty} \dfrac{x(\ln(1+ax+x^2)-\ln(1-ax+x^2))}{1+x^4}dx

Observe that, F(2)=IJF(\sqrt{2})=I-J and F(0)=0F(0)=0,

F(a)=0+2x(1+x2)(x2ax+1)(x2+ax+1)(x4+1)dxF'(a)= \int_0^{+\infty} \dfrac{2x(1+x^2)}{(x^2-ax+1)(x^2+ax+1)(x^4+1)}dx

F(a)=[2(arctan(2xa4a2)+arctan(2x+a4a2))4a2(a22)2(arctan(2x1)+arctan(2x+1))a22]0+F'(a)=\left[\dfrac{2 \left( \mathrm{arctan}\left( \dfrac{2x-a}{\sqrt{4-{{a}^{2}}}}\right) +\mathrm{arctan}\left( \dfrac{2x+a}{\sqrt{4-{{a}^{2}}}}\right) \right) }{\sqrt{4-{{a}^{2}}}\cdot \left( {{a}^{2}}-2\right) }-\dfrac{\sqrt{2} \left( \mathrm{arctan}\left( \sqrt{2}x-1\right) +\mathrm{arctan}\left( \sqrt{2}x+1\right) \right) }{{{a}^{2}}-2}\right]_0^{+\infty}

Therefore,

F(a)=2π(a22)4a22πa22F'(a)=\dfrac{2\pi}{(a^2-2)\sqrt{4-a^2}}-\dfrac{\sqrt{2}\pi}{a^2-2}

F(2)=02F(a)da=π2[ln((2+a)(4a2a)(2a)(4a2+a))]02=πln22F(\sqrt{2})=\int_0^{\sqrt{2}} F'(a)da=\dfrac{\pi}{2}\left[\mathrm{ln}\left( \dfrac{\left( \sqrt{2}+a\right)\left( \sqrt{4-{{a}^{2}}}-a\right) }{\left( \sqrt{2}-a\right)\left( \sqrt{4-{{a}^{2}}}+a\right) }\right)\right]_0^{\sqrt{2}}=\dfrac{\pi\ln 2}{2}

Therefore,

I+J=IJ=πln22I+J=I-J=\dfrac{\pi\ln 2}{2}

and,

I=πln22\boxed{I=\dfrac{\pi\ln 2}{2}} J=0J=0

FDP DPF - 4 years, 5 months ago

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You could speed up your calculation of I+JI+J by doing a Beta function trick. It is equal to 14ddaB(12,a12)  a=1 -\tfrac14 \frac{d}{da} B(\tfrac12,a-\tfrac12)\; \Big|_{a=1}

Mark Hennings - 4 years, 5 months ago

I came up with a similar solution just now, and was about to post, but you beat me to it by 4 hours!

Ishan Singh - 4 years, 5 months ago

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@Ishan Singh This one is a classic one ;)

FDP DPF - 4 years, 5 months ago

Problem 32 : Prove That

0(1x21+x2)ln(x)arctan(xx2+1) dxx=π320 \int_{0}^{\infty} \left( \dfrac{1-x^2}{1+x^2} \right) \cdot \ln (x) \cdot \arctan \left( \dfrac{x}{x^2 + 1} \right) \ \dfrac{\mathrm{d}x}{x} = - \dfrac{\pi^3}{20}

This problem has been solved by Mark Hennings.

Ishan Singh - 4 years, 5 months ago

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Suppose that u>0u > 0. Integrating by parts, we see note that 0tsechttanht1+14u2sech2tdt=[2tutan1(12usecht)]0+2u0tan1(12usecht)dt=2u0tan1(12usecht)dt\begin{aligned} \int_0^\infty \frac{t \,\mathrm{sech}\, t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt & = \Big[-\frac{2t}{u}\tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\Big]_0^\infty + \frac{2}{u}\int_0^\infty \tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\,dt \\ & = \frac{2}{u}\int_0^\infty \tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\,dt \end{aligned} Now the substitution v=12usechtv = \tfrac12u\mathrm{sech}\, t yields 0tsechttanht1+14u2sech2tdt=2u12u0tan1v(u)vu24v2dv=2012utan1vdvvu24v2  =  2u01tan1(12uw)w1w2dw=πuln(1+14u2+12u)\begin{aligned} \int_0^\infty \frac{t \,\mathrm{sech}\, t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt & = \frac{2}{u}\int_{\frac12u}^0 \frac{\tan^{-1}v\, (-u)}{v\sqrt{u^2-4v^2}}\,dv \\ & = 2\int_0^{\frac12u} \frac{\tan^{-1}v\,dv}{v\sqrt{u^2-4v^2}} \; = \; \frac{2}{u}\int_0^1 \frac{\tan^{-1}(\frac12 u w)}{w\sqrt{1-w^2}}\,dw \\ & = \frac{\pi}{u}\ln\Big(\sqrt{1 + \tfrac14u^2} +\tfrac12u\Big) \end{aligned} Note that the identity 01tan1pww1w2dw  =  12πln(1+p2+p) \int_0^1 \frac{\tan^{-1} pw}{w\sqrt{1-w^2}}\,dw \; = \; \tfrac12\pi\ln\big(\sqrt{1 + p^2} + p\big) is a standard integral from Gradshteyn & Ryzhik at the last stage. Thus, using the substitution x=etx = e^t, 01x21+x2lnxtan1(xx2+1)dxx=Rttanhttan1(12secht)dt=20ttanhttan1(12secht)dt=20ttanht(0112sechtdu1+14u2sech2t)dt=01(0tsechttanht1+14u2sech2tdt)du=π01ln(1+14u2+12u)duu  =  120π3\begin{aligned} \int_0^\infty \frac{1-x^2}{1+x^2} \ln x \tan^{-1}\left(\frac{x}{x^2+1}\right)\,\frac{dx}{x} & = - \int_{\mathbb{R}}t \tanh t \tan^{-1}\big(\tfrac12\,\mathrm{sech}\,t\big)\,dt \\ & = -2\int_0^\infty t \tanh t \tan^{-1}\big(\tfrac12\,\mathrm{sech}\,t\big)\,dt \\ & = -2\int_0^\infty t \tanh t \left(\int_0^1 \frac{\frac12\,\mathrm{sech}\,t\,du}{1 + \frac14u^2\,\mathrm{sech}^2t}\right)\,dt \\ & = -\int_0^1\left(\int_0^\infty \frac{t\,\mathrm{sech}\,t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt\right)\,du \\ & = -\pi\int_0^1 \ln\Big(\sqrt{1 + \tfrac14u^2} + \tfrac12u\Big)\,\frac{du}{u} \; = \; -\tfrac{1}{20}\pi^3 \end{aligned} using the result of Problem 30.

Mark Hennings - 4 years, 5 months ago

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Nice. By the way, Taylor expansion of 1xarctan(x1+x2)\dfrac{1}{x}\arctan\left(\dfrac{x}{1+x^2}\right) is n=0+(1)nL2n+1x2n2n+1\displaystyle \sum_{n=0}^{+\infty} \dfrac{(-1)^n L_{2n+1}x^{2n}}{2n+1}

and,

Taylor expansion of arctan(x1+x2)x(1+x2)\dfrac{\arctan\left(\dfrac{x}{1+x^2}\right)}{x(1+x^2)} is n=0+(1)n(p=0nL2p+12p+1)x2n\displaystyle \sum_{n=0}^{+\infty} (-1)^n\left(\sum_{p=0}^n \dfrac{L_{2p+1}}{2p+1}\right)x^{2n}

LkL_k is the kk-th Lucas number.

FDP DPF - 4 years, 5 months ago

(+1) Nice! My method was a bit different (used tangent half angle in the beginning), but I also reduce it to P30. The integral you mentioned can be proved using differentiation under the integral and forming a differential equation.

Ishan Singh - 4 years, 5 months ago

Problem 34:

Prove that 11+x2+x4++x2kdx=2πk+1cos(π2k+2)sin(3π2k+2)  . \large\int_{-\infty}^\infty \dfrac1{1+x^2+x^4+\cdots+x^{2k} } \, dx = \dfrac{2\pi}{k+1} \cdot \dfrac{ \cos\left( \frac{\pi}{2k+2} \right) }{ \sin \left( \frac{3\pi}{2k+2} \right) } \; .

This problem has been solved by Mark Hennings.

Pi Han Goh - 4 years, 5 months ago

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If we define ζ=eπik+1\zeta = e^{\frac{\pi i}{k+1}}, then fk(z)  =  j=0kzj  =  zk+11z1  =  j=1k(zζ2j) f_k(z) \; = \; \sum_{j=0}^k z^j \; = \; \frac{z^{k+1}-1}{z-1} \; = \; \prod_{j=1}^k \big(z - \zeta^{2j}\big) and hence, using partial fractions, we must be able to write 1fk(z)  =  j=1kAjzζ2j \frac{1}{f_k(z)} \; = \; \sum_{j=1}^k \frac{A_j}{z - \zeta^{2j}} where Aj  =  limzζ2jzζ2jfk(z)  =  [fk(ζ2j)]1  =  ζ2j(ζ2j1)k+1 A_j \; = \; \lim_{z \to \zeta^{2j}} \frac{z - \zeta^{2j}}{f_k(z)} \; = \; \big[f_k'\big(\zeta^{2j}\big)\big]^{-1} \; = \; \frac{\zeta^{2j}(\zeta^{2j}-1)}{k+1} Thus 1fk(z2)  =  j=1kAjz2ζ2j  =  j=1kAj2ζj(1zζj1z+ζj) \frac{1}{f_k(z^2)} \; = \; \sum_{j=1}^k \frac{A_j}{z^2 - \zeta^{2j}} \; =\; \sum_{j=1}^k \frac{A_j}{2\zeta^j}\left(\frac{1}{z - \zeta^j} - \frac{1}{z + \zeta^j}\right) and hence the function 1fk(z2) \frac{1}{f_k(z^2)} has simple poles at ±ζj\pm \zeta^j for 1jk1 \le j \le k. Thus 1fk(x2)dx=2πij=1kResz=ζj1fk(z2)  =  2πij=1kAj2ζj=πik+1j=1kζj(ζ2j1)  =  πik+1[ζ3(ζ3k1)ζ31ζ(ζk1)ζ1]=πik+1[ζ3(ζ31)ζ31ζ(ζ11)ζ1]  =  πik+1[ζ+1ζ1ζ3+1ζ31]=2πik+1×ζ(ζ21)(ζ1)(ζ31)  =  2πik+1×ζ(ζ+1)ζ31=2πik+1×e3πi2(k+1)2cos(π2(k+1))e3πi2(k+1)2isin(3π2(k+1))=2πk+1cos(π2(k+1))sin(3π2(k+1))\begin{aligned} \int_{-\infty}^\infty \frac{1}{f_k(x^2)}\,dx & = 2\pi i\sum_{j=1}^k \mathrm{Res}_{z = \zeta^j} \frac{1}{f_k(z^2)} \; = \; 2\pi i \sum_{j=1}^k \frac{A_j}{2\zeta^j} \\ & = \frac{\pi i}{k+1}\sum_{j=1}^k \zeta^j(\zeta^{2j} - 1) \; = \; \frac{\pi i}{k+1}\left[ \frac{\zeta^3(\zeta^{3k} - 1)}{\zeta^3 - 1} - \frac{\zeta(\zeta^k - 1)}{\zeta-1}\right] \\ & = \frac{\pi i}{k+1} \left[\frac{\zeta^3(-\zeta^{-3} - 1)}{\zeta^3-1} - \frac{\zeta(\zeta^{-1}-1)}{\zeta-1}\right] \; = \; \frac{\pi i}{k+1}\left[\frac{\zeta+1}{\zeta-1} - \frac{\zeta^3+1}{\zeta^3-1}\right] \\ & = \frac{2 \pi i}{k+1} \times \frac{\zeta(\zeta^2-1)}{(\zeta-1)(\zeta^3-1)} \; = \; \frac{2\pi i}{k+1} \times \frac{\zeta(\zeta+1)}{\zeta^3-1} \\ & = \frac{2\pi i}{k+1} \times \frac{e^{\frac{3\pi i}{2(k+1)}} 2\cos\big(\frac{\pi}{2(k+1)}\big)}{e^{\frac{3\pi i}{2(k+1)}} 2i \sin\big(\frac{3\pi}{2(k+1)}\big)} \\ & = \frac{2\pi}{k+1}\,\frac{\cos\big(\frac{\pi}{2(k+1)}\big)}{\sin\big(\frac{3\pi}{2(k+1)}\big)} \end{aligned} as required.

Mark Hennings - 4 years, 5 months ago

Problem 35:

Show that

01lnxx2+x+1dx  =  29[23π2ψ(13)] \int_0^1 \frac{\ln x}{x^2 + x + 1}\,dx \; = \; \tfrac29\left[\tfrac23\pi^2 - \psi'(\tfrac13)\right]

This problem has been solved by Fdp Dpf.

Mark Hennings - 4 years, 5 months ago

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01lnx1+x+x2dx=011x1x3lnxdx=0111x3lnxdx01x1x3lnxdx=n=0+(01x3nlnx)n=0+(01x3n+1lnx)=n=0+1(3n+1)2+n=0+1(3n+2)2=n=0+1(3n+1)2+ζ(2)n=0+1(3n+1)2n=1+1(3n)2=2n=0+1(3n+1)2+89ζ(2)=427π229n=0+1(n+13)2=427π229ψ(13)\begin{aligned}\int_0^1 \dfrac{\ln x}{1+x+x^2}dx&=\int_0^1 \dfrac{1-x}{1-x^3}\ln x dx\\ &=\int_0^1 \dfrac{1}{1-x^3}\ln x dx-\int_0^1 \dfrac{x}{1-x^3}\ln x dx\\ &=\sum_{n=0}^{+\infty}\left(\int_0^1 x^{3n}\ln x\right)-\sum_{n=0}^{+\infty}\left(\int_0^1 x^{3n+1}\ln x\right)\\ &=-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\sum_{n=0}^{+\infty} \dfrac{1}{(3n+2)^2}\\ &=-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\zeta(2)-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}-\sum_{n=1}^{+\infty} \dfrac{1}{(3n)^2}\\ &=-2\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\dfrac{8}{9}\zeta(2)\\ &=\dfrac{4}{27}\pi^2 -\dfrac{2}{9}\sum_{n=0}^{+\infty} \dfrac{1}{\left(n+\tfrac{1}{3}\right)^2}\\ &=\boxed{\dfrac{4}{27}\pi^2 -\dfrac{2}{9}\psi^\prime \left(\dfrac{1}{3}\right)} \end{aligned} Since ψ(a)=n=0+1(n+a)2\displaystyle \psi^\prime \left(a\right)=\sum_{n=0}^{+\infty} \dfrac{1}{(n+a)^2}

FDP DPF - 4 years, 5 months ago

PROBLEM 39:

Evaluate 0(exa11+xb)dxx \int_0^\infty\left(e^{-x^a} - \frac{1}{1+x^b}\right)\,\frac{dx}{x} for any a,b>0a,b > 0.

Mark Hennings - 4 years, 5 months ago

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Solution to problem 39.

J=0+(exa11+xb)1xdx J=\int_0^{+\infty} \left( \text{e}^{-x^a}-\dfrac{1}{1+x^b}\right)\dfrac{1}{x}dx

Perform the change of variable y=xay=x^a,

J=1a0+(ex11+xab)1xdxJ=\dfrac{1}{a}\int_0^{+\infty} \left( \text{e}^{-x}-\dfrac{1}{1+x^{\tfrac{a}{b}}}\right)\dfrac{1}{x}dx

Perform integration by parts,

aJ=[(ex11+xab)lnx]0++0+(exabxab1(1+xab)2)lnxdx=0+exlnxdxab0+xab1(1+xab)2lnxdx\begin{aligned}aJ&=\left[ \left( \text{e}^{-x}-\dfrac{1}{1+x^{\tfrac{a}{b}}}\right)\ln x\right]_0^{+\infty}+\int_0^{+\infty}\left( \text{e}^{-x}-\dfrac{a}{b}\dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\right)\ln x dx\\ &=\int_0^{+\infty} \text{e}^{-x}\ln x dx-\dfrac{a}{b} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx \end{aligned}

0+xab1(1+xab)2lnxdx=01xab1(1+xab)2lnxdx+1+xab1(1+xab)2lnxdx\begin{aligned} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx&=\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx+\int_1^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx \end{aligned}

In the latter integral perform the change of variable y=1xy=\dfrac{1}{x},

0+xab1(1+xab)2lnxdx=01xab1(1+xab)2lnxdx01xab1(1+xab)2lnxdx=0\begin{aligned} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx&=\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx-\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx\\ &=0 \end{aligned}

0+exlnxdx=Γ(1)=γ\int_0^{+\infty} \text{e}^{-x}\ln x dx=\Gamma^\prime (1)=-\gamma

Therefore,

J=1aγ\boxed{J=-\dfrac{1}{a}\gamma}

FDP DPF - 4 years, 5 months ago

Problem 43 : Prove That 0dx1+(x+tanx)2=π2 \int_{0}^{\infty} \dfrac{\mathrm{d}x}{1 + (x+\tan x)^2} = \dfrac{\pi}{2}

Ishan Singh - 4 years, 5 months ago

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Note that,

tanx=limn(1xπ2+1x+π2++1x(2n1)π2+1x+(2n1)π2) \tan x = - \lim_{n \to \infty} \left( \dfrac{1}{x- \frac{\pi}{2}} + \dfrac{1}{x+ \frac{\pi}{2}} + \ldots + \dfrac{1}{x- (2n-1)\frac{\pi}{2}} + \dfrac{1}{x + (2n-1)\frac{\pi}{2}} \right)

Also, by Glasser's Master Theorem (which can be proved using elementary methods by using the graphical properties of the function), we have,

+f(xa1xλ1anxλn)dx=+f(x)dx;ai>0 , λiR \int_{-\infty}^{+\infty}f\left(x-\frac{a_1}{x-\lambda_1}-\cdots-\frac{a_n}{x-\lambda_n}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \quad ; \quad a_{i} >0 \ , \ \lambda_{i} \in \mathbb{R}

Therefore,

+f(x+atanx)dx=+f(x)dx;a>0 \int_{-\infty}^{+\infty}f\left(x+ a\tan x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \quad ; \quad a >0

Putting f(x)=1x2+b2f(x) = \dfrac{1}{x^2+b^2} , we have,

dx(x+atanx)2+b2=dxx2+b2 \int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{(x+a \tan x)^2+b^2} = \int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{x^2+b^2}

0dx(x+atanx)2+b2=π2b \therefore \int_{0}^{\infty} \dfrac{\mathrm{d}x}{(x+a \tan x)^2+b^2} = \dfrac{\pi}{2b} \quad \square

Ishan Singh - 4 years, 5 months ago

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I think there are some big convergence issues to be settled The convergence of the series for tanx\tan x is anything but uniform, and there is no function that I can see that will enable the DCT to guarantee to obtain f(x+atanx)dx \int_{-\infty}^\infty f(x+a\tan x)\,dx as the limit of f(xa1xλ1anxλn)dx \int_{-\infty}^\infty f\left(x - \tfrac{a_1}{x-\lambda_1} - \cdots - \tfrac{a_n}{x - \lambda_n}\right)\,dx

Mark Hennings - 4 years, 5 months ago

Here is my take, using the ideas I had been playing with, which convert the integral into a shape for which the GMT can be used, and a limit taken, to get the answer.

Write I  =  011+(x+tanx)2dx=12Rdx1+(x+tanx)2=12nZ12π12π11+(nπ+x+tanx)2dx=12π212π12πnZdx(n+x+tanxπ)2+1π2=12π212π12ππ2sinh2cosh2cos(2x+2tanx)dx=12sinh212π12πdxcosh2cos(2x+2tanx)\begin{aligned} I \; = \; \int_0^\infty \frac{1}{1 + (x + \tan x)^2}\,dx & = \tfrac12\int_{\mathbb{R}} \frac{dx}{1 + (x + \tan x)^2} \\ & = \tfrac12 \sum_{n \in \mathbb{Z}} \int_{-\frac12\pi}^{\frac12\pi} \frac{1}{1 + (n\pi + x + \tan x)^2}\,dx \\ & = \tfrac{1}{2\pi^2}\int_{-\frac12\pi}^{\frac12\pi} \sum_{n \in \mathbb{Z}} \frac{dx}{\big(n + \tfrac{x + \tan x}{\pi}\big)^2 + \tfrac{1}{\pi^2}} \\ & = \frac{1}{2\pi^2} \int_{-\frac12\pi}^{\frac12\pi} \frac{\pi^2 \sinh 2}{\cosh 2 - \cos(2x + 2\tan x)}\,dx \\ & = \tfrac12\sinh2 \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{\cosh 2 - \cos(2x + 2\tan x)} \end{aligned} Since we are integrating over a finite interval, and since the integrand is bounded on the interval, we can use the GMT, taking the limit in the manner you suggest, so that I  =  12sinh212π12πdxcosh2cos2x  =  14sinh2ππdxcosh2cosx I \; = \; \tfrac12\sinh2 \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{\cosh2 - \cos 2x} \; = \; \tfrac14\sinh2 \int_{-\pi}^\pi \frac{dx}{\cosh 2 - \cos x} Making the complex substitution z=eixz = e^{ix}, this becomes I=14sinh2z=11cosh212(z+z1)dziz  =  sinh22iz=1dz(ze2)(ze2)=πsinh2Resz=e21(ze2)(ze2)  =  12π\begin{aligned} I & = \tfrac14\sinh2 \int_{|z|=1} \frac{1}{\cosh2 - \frac12(z + z^{-1})} \frac{dz}{iz} \; = \; -\frac{\sinh 2}{2i}\int_{|z|=1}\frac{dz}{(z-e^2)(z-e^{-2})} \\ & = -\pi \sinh 2\, \mathrm{Res}_{z = e^{-2}} \frac{1}{(z - e^2)(z - e^{-2})} \; = \; \tfrac12\pi \end{aligned} as required.

Mark Hennings - 4 years, 5 months ago

Problem 38:

Evaluate,

0π4xln(cosx+sinxcosxsinx)cosx(cosx+sinx)dx\displaystyle \int_0^{\tfrac{\pi}{4}} \dfrac{x\ln\left(\tfrac{\cos x+\sin x}{\cos x-\sin x}\right)}{\cos x(\cos x+\sin x)}\, dx

FDP DPF - 4 years, 5 months ago

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We note that I=014πxcosx(cosx+sinx)ln(cosx+sinxcosxsinx)dx  =  014πx1+tanxln(1+tanx1tanx)sec2xdx=01tan1u1+uln(1+u1u)du  =  01tan1(1y1+y)lny1+ydy\begin{aligned} I & = \int_0^{\frac14\pi}\frac{x}{\cos x(\cos x + \sin x)} \ln\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right)\,dx \; = \; \int_0^{\frac14\pi} \frac{x}{1 + \tan x}\ln\left(\frac{1 + \tan x}{1 - \tan x}\right)\,\sec^2x\,dx \\ & = \int_0^1 \frac{\tan^{-1}u}{1+u}\ln\left(\frac{1+u}{1-u}\right)\,du \; = \; -\int_0^1 \tan^{-1}\left(\frac{1-y}{1+y}\right)\,\frac{\ln y}{1+y}\,dy \end{aligned} after the substitutions u=tanxu = \tan x and y=1u1+uy = \frac{1-u}{1+u}. Since tan(14πtan1y)  =  1y1+y \tan\big(\tfrac14\pi - \tan^{-1}y\big) \; = \; \frac{1-y}{1+y} we deduce that I  =  01(14πtan1y)lny1+ydy  =  14π01lny1+ydy+01tan1ylny1+ydy I \; = \; -\int_0^1 \left(\tfrac14\pi - \tan^{-1}y\right)\frac{\ln y}{1+y}\,dy \; = \; -\tfrac14\pi\int_0^1 \frac{\ln y}{1+y}\,dy + \int_0^1 \frac{\tan^{-1}y \ln y}{1+y}\,dy Now it is elementary that 01lny1+ydy  =  112π2 \int_0^1 \frac{\ln y}{1+y}\,dy \; = \; -\tfrac{1}{12}\pi^2 and we see from this MSE post --- hello, it's one of yours again --- that 01tan1ylny1+ydy  =  12Gln2164π3 \int_0^1 \frac{\tan^{-1}y \ln y}{1+y}\,dy \; = \; \tfrac12 G \ln 2 - \tfrac{1}{64}\pi^3 so it follows that I  =  1192π3+12Gln2 I \; = \; \tfrac{1}{192}\pi^3 + \tfrac12G \ln 2

Mark Hennings - 4 years, 5 months ago

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Congrats ! It's like you read my mind, you know all my tricks ;) Happy new year to all !

FDP DPF - 4 years, 5 months ago

Problem 40

Evaluate 0x12x1 ln(2x1)dx\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx

FDP DPF - 4 years, 5 months ago

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Substitute (2x1)=t2(2^x-1) = t^2 to get,

I=1ln220(ln(t2+1)ln2(t2+1)lnt)dt \text{I} = \displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t

Substitute t1tt \mapsto \dfrac{1}{t}

    I=1ln220(ln(t2+1)ln2(t2+1)lnt)dt+2ln220dtt2+1 \implies \text{I} = -\displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t + \dfrac{2}{\ln^2 2} \int_{0}^{\infty} \dfrac{\mathrm{d}t}{t^2+1}

    I=I+πln22 \implies \text{I} = -\text{I} + \dfrac{\pi}{\ln^2 2}

    I=π2ln22 \implies \text{I} = \dfrac{\pi}{2 \ln^2 2}

Ishan Singh - 4 years, 5 months ago

PROBLEM 42 :

Evaluate 01(xp1)(xq1)(x1)lnxdxp,q>0 \int_0^1 \frac{(x^p - 1)(x^q - 1)}{(x-1) \ln x}\,dx \hspace{2cm} p,q > 0

Mark Hennings - 4 years, 5 months ago

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Let,

I=01(xp1)(xq1)(x1)lnx dx\displaystyle \text{I} = \int_{0}^{1} \dfrac{(x^p - 1)(x^q-1)}{(x-1) \ln x} \ \mathrm{d}x

=01(xp1)(xq1)(1x)lnx dx\displaystyle = - \int_{0}^{1} \dfrac{(x^p - 1)(x^q-1)}{(1-x) \ln x} \ \mathrm{d}x

=r=001[(xp+q+rxp+rlnx)(xq+rxrlnx)] dx\displaystyle = - \sum_{r=0}^{\infty} \int_{0}^{1} \left[ \left(\dfrac{x^{p+q+r} - x^{p+r}}{\ln x}\right) - \left(\dfrac{x^{q+r} - x^{r}}{\ln x}\right) \right] \ \mathrm{d}x

=limnr=0n[log(p+q+r+1p+r+1)log(q+r+1r+1)]\displaystyle = -\lim_{n \to \infty} \sum_{r=0}^{n} \left[\log \left(\dfrac{p+q+r+1}{p+r+1}\right) - \log\left( \dfrac{q+r+1}{r+1} \right)\right]

=limnr=1n[log(p+q+rp+r)log(q+rr)]\displaystyle = -\lim_{n \to \infty} \sum_{r=1}^{n} \left[\log \left(\dfrac{p+q+r}{p+r}\right) - \log\left( \dfrac{q+r}{r} \right)\right]

where the standard result 01xaxblnx dx=ln(a+1b+1)\displaystyle \int_{0}^{1}\dfrac{x^a-x^b}{\ln x} \ \mathrm{d}x = \ln\left( \dfrac{a+1}{b+1} \right) has been used in the above lines.

    I=limnlog(n!×(p+q+1)(p+q+2)(p+q+n)[(p+1)(p+2)(p+n)]×[(q+1)(q+2)(q+n)])\displaystyle \implies \text{I} = -\lim_{n \to \infty} \log \left( \dfrac{n! \times (p+q+1)(p+q+2)\ldots(p+q+n)}{[(p+1)(p+2)\ldots(p+n)] \times [(q+1)(q+2)\ldots(q+n)]} \right)

=limnlog(Γ(p+q+n+1)Γ(p+1)Γ(n+1)Γ(q+1)Γ(p+q+1)Γ(p+n+1)Γ(q+n+1)) \displaystyle = -\lim_{n \to \infty} \log \left( \dfrac{\Gamma(p+q+n+1) \Gamma(p+1) \Gamma(n+1) \Gamma(q+1) }{\Gamma(p+q+1) \Gamma(p+n+1) \Gamma(q+n+1)} \right)

Since limnnxΓ(n+1)Γ(n+x+1)=1\displaystyle \lim_{n\to \infty} \dfrac{n^x \Gamma(n+1)}{\Gamma(n+x+1)} = 1 , we have,

I=log(Γ(p+q+1)Γ(p+1)Γ(q+1)) \displaystyle \text{I} = \log\left( \dfrac{\Gamma(p+q+1)}{\Gamma(p+1) \Gamma(q+1)} \right) \quad \square

Ishan Singh - 4 years, 5 months ago

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One of the standard integral representations of lnΓ(z)\ln \Gamma(z) gets you there much more easily...

Mark Hennings - 4 years, 5 months ago

Problem 33:

Evaluate 014πsinpxcosp+2xdxp>0  .\large \int_0^{\frac 14\pi} \frac{\sin^px}{\cos^{p+2}x}\,dx \hspace{2cm} p > 0 \;.

This problem has been solved by Sumanth R Hegde.

Mark Hennings - 4 years, 5 months ago

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Put tanx=ttanx = t

The integral reduces to 01tpdt \int\limits_{0}^{1} t^p dt

This gives 1p+1\frac{1}{p+1}

Anyone else may post the next question

Sumanth R Hegde - 4 years, 5 months ago

Problem 41 :

Prove That

(nx) dx=2n(n)>1 \int_{-\infty}^{\infty} \dbinom{n}{x} \ \mathrm{d}x = 2^n \quad \Re(n) > -1

Notation : (nx) \dbinom{n}{x} denotes the Generalized Binomial Coefficient.

Ishan Singh - 4 years, 5 months ago

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Presumably, we mean (nx)  =  n!Γ(x+1)Γ(n+1x)xR {n \choose x} \; = \; \frac{n!}{\Gamma(x+1)\Gamma(n+1-x)} \hspace{2cm} x \in \mathbb{R} Now Γ(x+1)Γ(n+1x)  =  xΓ(x)j=1n(jx)Γ(1x)  =  (1)nπsinπx×j=0n(xj) \Gamma(x+1)\Gamma(n+1-x) \; = \; x\Gamma(x) \prod_{j=1}^n (j-x)\Gamma(1-x) \; = \; (-1)^n \frac{\pi}{\sin\pi x}\times \prod_{j=0}^n (x-j) and so we can use partial fractions to write (nx)  =  (1)nn!πsinπx(j=0n(xj))1  =  (1)nn!sinπxπj=0nAjxj {n \choose x} \; = \; \frac{(-1)^n n!}{\pi} \sin \pi x \left(\prod_{j=0}^n (x-j)\right)^{-1} \; = \; \frac{(-1)^n n! \sin \pi x}{\pi} \sum_{j=0}^n \frac{A_j}{x-j} where Aj  =  (1)njj!(nj)!0jn A_j \; = \; \frac{(-1)^{n-j}}{j! (n-j)!} \hspace{1cm} 0 \le j \le n Since Rsinπxxjdx  =  (1)jRsinπxxdx  =  (1)jπ \int_{\mathbb{R}} \frac{\sin \pi x}{x - j}\,dx \; = \; (-1)^j \int_{\mathbb{R}} \frac{\sin \pi x}{x}\,dx \; = \; (-1)^j \pi we deduce that R(nx)dx=(1)nn!πj=0nAjRsinπxxjdx  =  (1)nn!j=0n(1)jAj=j=0nn!j!(nj)!  =  j=0n(nj)  =  2n\begin{aligned} \int_{\mathbb{R}} {n \choose x}\,dx & = \frac{(-1)^n n!}{\pi} \sum_{j=0}^n A_j \int_{\mathbb{R}}\frac{\sin \pi x}{x-j}\,dx \; = \; (-1)^n n! \sum_{j=0}^n (-1)^j A_j \\ & = \sum_{j=0}^n \frac{n!}{j!(n-j)!} \; = \; \sum_{j=0}^n {n \choose j} \; = \; 2^n \end{aligned} as required.

If we insist on being picky, we could integrate from X-X to XX, and let XX \to \infty, to avoid dealing with the improper integral Rsinxxdx \int_{\mathbb{R}} \frac{\sin x}{x}\,dx directly. We only have to combine a finite number of integrals at the last stage, so there is no problem taking the limit.

Mark Hennings - 4 years, 5 months ago

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The integral is also true for (n)>1\Re(n) > -1.

Ishan Singh - 4 years, 5 months ago

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@Ishan Singh Indeed. The formulae dxΓ(α+x)Γ(βx)  =  2α+β2Γ(α+β1) \int_{-\infty}^\infty \frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)} \; = \; \frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)} for Re(α+β)>1\mathfrak{Re}(\alpha+\beta) > 1 is 6.414.26.414.2 in Gradshteyn & Rhyzik.

Isn't it a bit late to be altering the question, though?

Mark Hennings - 4 years, 5 months ago

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@Mark Hennings Apologies, I forgot to mention that initially. But your answer is the accepted one nevertheless.

Ishan Singh - 4 years, 5 months ago

I think your solution can be extended to deal with non integral n if we instead expand into partial fractions using infinite product of Gamma Function. My approach was using the integral representation 0π(sin(x))m1einx dx=π2m1 ein/2mB(12(m+n+1),12(mn+1))\large \int_{0}^{\pi} (\sin (x))^{m-1} e^{inx} \ \mathrm{d}x = \dfrac{\pi}{2^{m-1}} \ \dfrac{e^{i n/2}}{m \operatorname{B} \left( \dfrac{1}{2} (m+n+1) , \dfrac{1}{2} (m-n+1) \right)}

Ishan Singh - 4 years, 5 months ago

If (n+1x+1)=(nx)+(nx+1)\binom{n+1}{x+1}=\binom{n}{x}+\binom{n}{x+1} then i think it's straightforward to prove with a recurrence proof.

FDP DPF - 4 years, 5 months ago

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Certainly, if we presume that the integral exists, then the recurrence relation can help us evaluate it. The RR does not prove convergence of the integral, however.

Mark Hennings - 4 years, 5 months ago

Problem 45.

Evaluate 0x5sinh(πx)(1+x6)dx\int_0^\infty \frac{x^5}{\sinh(\pi x)(1+x^6)}dx

FDP DPF - 4 years, 5 months ago

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Using the Fourier transform we have I  =  0x5sinhπx(1+x6)dx  =  12x5sinhπx(1+x6)dx  =  12f,g  =  12Ff,Fg I \; = \; \int_0^\infty \frac{x^5}{\sinh \pi x(1 + x^6)}\,dx \; =\; \tfrac12\int_{-\infty}^\infty \frac{x^5}{\sinh \pi x(1 + x^6)}\,dx \; = \; \tfrac12\langle f\,,\,g\rangle \; = \; \tfrac12\langle \mathcal{F}f\,,\,\mathcal{F}g \rangle where f(x)  =  xsinhπxg(x)  =  x41+x6 f(x) \; = \; \frac{x}{\sinh \pi x} \hspace{2cm} g(x) \; = \; \frac{x^4}{1 + x^6} A standard result about Fourier transforms tells us that (Ff)(u)  =  122πsech212u (\mathcal{F}f)(u) \; = \; \frac{1}{2\sqrt{2\pi}}\,\mathrm{sech}^2\tfrac12u and the Fourier transform of gg can be readily calculated using contour integration: for u>0u > 0 we simply need to determine the residues of z4eizu1+z6\frac{z^4e^{izu}}{1+z^6} at the three roots of z6+1=0z^6 + 1 = 0 with positive imaginary part. Omitting the details, (Fg)(u)  =  13π2[eu+e12ucos(32u)3e12usin(32u)] (\mathcal{F}g)(u) \; = \; \tfrac13\sqrt{\tfrac{\pi}{2}}\Big[e^{-|u|} + e^{-\frac12|u|}\cos\big(\tfrac{\sqrt{3}}{2}|u|\big) - \sqrt{3}e^{-\frac12|u|}\sin\big(\tfrac{\sqrt{3}}{2}|u|\big)\Big] Thus we deduce that I  =  12Ff,Fg  =  1120sech212u[eu+e12ucos(32u)3e12usin(32u)]du I \; = \; \tfrac12\langle \mathcal{F}f\,,\,\mathcal{F}g\rangle \; = \; \tfrac{1}{12}\int_0^\infty \mathrm{sech}^2\tfrac12u\,\Big[e^{-u} + e^{-\frac12u}\cos\big(\tfrac{\sqrt{3}}{2}u\big) - \sqrt{3}e^{-\frac12u}\sin\big(\tfrac{\sqrt{3}}{2}u\big)\Big]\,du Now 1va(1+v)2du  =  12(1aH12a2+aH12a)Rea<1 \int_1^\infty \frac{v^a}{(1+v)^2}\,du \; = \; \tfrac12\Big(1 - aH_{-\frac12-\frac{a}{2}} + a H_{-\frac12a}\Big) \hspace{2cm} \mathfrak{Re}\,a < 1 and so 0eωusech212udu  =  40eωu(e12u+e12u)2du  =  41vω(1+v)2dv  =  2(1ωH12ω2+ωH12ω) \int_0^\infty e^{\omega u}\,\mathrm{sech}^2\tfrac12u\,du \; = \; 4\int_0^\infty \frac{e^{\omega u}}{(e^{\frac12u} + e^{-\frac12u})^2}\,du \; = \; 4\int_1^\infty \frac{v^\omega}{(1 + v)^2}\,dv \; = \; 2\Big(1 - \omega H_{\frac12\omega^2} + \omega H_{-\frac12\omega}\Big) where ω\omega is the primitive cube root of unity. Taking real and imaginary parts appropriately, and doing a lot of simplification with polygammas, 0e12u[cos(32u)3sin(32u)]sech212udu  =  2+4πsech(3π2) \int_0^\infty e^{-\frac12u}\Big[\cos\big(\tfrac{\sqrt{3}}{2}u\big) - \sqrt{3}\sin\big(\tfrac{\sqrt{3}}{2}u\big)\Big]\,\mathrm{sech}^2\tfrac12u\,du \; = \; -2 + 4\pi\,\mathrm{sech}\big(\tfrac{\sqrt{3}\pi}{2}\big) On the other hand, 0eusech212udu  =  41dvv(v+1)2  =  2+4ln2 \int_0^\infty e^{-u}\,\mathrm{sech}^2\tfrac12u\,du \; = \;4\int_1^\infty \frac{dv}{v(v+1)^2} \; = \; -2 + 4\ln 2 Putting this all together, we obtain that I  =  13[1+ln2+πsech(3π2)] I \; = \; \tfrac13\Big[-1 + \ln2 + \pi\,\mathrm{sech}\big(\tfrac{\sqrt{3}\pi}{2}\big)\Big] Someone else can post the next one, please.

Mark Hennings - 4 years, 5 months ago

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Here is a partial progress towards an alternate solution f(a)=0sinh(ax)sinh(πx)(1+x6) dx f(a) = \int_0^\infty \frac{\sinh (ax)}{\sinh(\pi x)(1+x^6)} \ \mathrm{d}x

then,

f(6)(a)+f(a)=12tan(a2)f^{(6)} (a) + f(a) = \dfrac{1}{2} \tan\left(\dfrac{a}{2}\right)

where f(n)(a)f^{(n)} (a) denotes the nthn^{\text{th}} derivative of ff w.r.t. aa, with f(0)=0f(0) = 0 and the required integral being f(5)(0)f^{(5)} (0).

I still have to solve that D.E.

Ishan Singh - 4 years, 5 months ago

* PROBLEM 46: *

Show that 011x2x4+x2+1dx  =  π(31)2234 \int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx \; = \; \frac{\pi(\sqrt{3}-1)}{2\sqrt{2}\sqrt[4]{3}}

Mark Hennings - 4 years, 5 months ago

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Perform the change of variable y=1x1+xy=\dfrac{1-x}{1+x},

011x2x4+x2+1dx=01x(4+4x)3x4+10x2+3dx=[(232334232354)atan(2y23142314)+(232334232354)atan(2314+2y2314)+(232334232354)atan(23y23142314)+(232334232354)atan(2314+23y2314)+(2354+2334)log(y2314y+3)+(23542334)log(y+2314y+3)+(23542334)log(3y2314y+1)+(2354+2334)log(3y+2314y+1)×124]01=(223)atan(3142314)+(232)atan(2+314314)+(232)atan(23141)+(232)atan(2314+1)4314=(62)(arctan(3+23)+πarctan(3+23))4×314=π(31)2234\begin{aligned} \int_0^1 \dfrac{\sqrt{1-x^2}}{x^4+x^2+1}dx&=\int_{0}^{1}\frac{\sqrt{x}\cdot \left( 4+4\cdot x\right) }{3\cdot {{x}^{4}}+10\cdot {{x}^{2}}+3}dx\\ &=\left[\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{2\cdot \sqrt{y}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}+2\cdot \sqrt{y}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right)+\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{2\cdot \sqrt{3}\cdot \sqrt{y}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}+2\cdot \sqrt{3}\cdot \sqrt{y}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot {{3}^{\frac{5}{4}}}+\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( y-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+\sqrt{3}\right) +\left( -\sqrt{2}\cdot {{3}^{\frac{5}{4}}}-\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( y+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+\sqrt{3}\right) +\left( -\sqrt{2}\cdot {{3}^{\frac{5}{4}}}-\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( \sqrt{3}\cdot y-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+1\right) +\left( \sqrt{2}\cdot {{3}^{\frac{5}{4}}}+\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( \sqrt{3}\cdot y+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+1\right)\times -\dfrac{1}{24}\right]_0^1\\ &=\frac{\left( \sqrt{2}-\sqrt{2}\cdot \sqrt{3}\right) \cdot \mathrm{atan}\left( \frac{{{3}^{\frac{1}{4}}}-\sqrt{2}}{{{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}+{{3}^{\frac{1}{4}}}}{{{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \sqrt{2}\cdot {{3}^{\frac{1}{4}}}-1\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \sqrt{2}\cdot {{3}^{\frac{1}{4}}}+1\right) }{4\cdot {{3}^{\frac{1}{4}}}}\\ &=\dfrac{\left(\sqrt{6}-\sqrt{2}\right)\left(\arctan\left(\sqrt{3+2\sqrt{3}}\right)+\pi-\arctan\left(\sqrt{3+2\sqrt{3}}\right)\right)}{4\times 3^{\tfrac{1}{4}}}\\ &=\boxed{\dfrac{\pi \left(\sqrt{3}-1\right)}{2\sqrt{2}\sqrt[4]{3}}} \end{aligned}

FDP DPF - 4 years, 5 months ago

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My approach used contour integration...

The function z1\sqrt{z - 1} can be defined analytically on the cut plane C\{(,1]\mathbb{C} \backslash \{(-\infty,1], while the function z+1\sqrt{z+1} can be defined analytically on the cut plane C\(,1]\mathbb{C} \backslash (-\infty,-1]. Putting these two functions together, it is possible to define the function z21\sqrt{z^2-1} analytically on the domain C\[1,1]\mathbb{C} \backslash [-1,1].

Consider the "dogbone" contour Dε=γ1+γ2+γ3+γ4D_\varepsilon = -\gamma_1 + \gamma_2 + \gamma_3 + \gamma_4, where

  • γ1\gamma_1 is the straight line segment z=xz = x for 1+ε<x<1ε-1+\varepsilon < x < 1-\varepsilon, just above the cut,
  • γ2\gamma_2 is the (almost) circular arc z=1+εeiθz = -1 + \varepsilon e^{i\theta} for 0<θ<2π0 < \theta < 2\pi,
  • γ3\gamma_3 is the straight line segment z=xz = x for 1+ε<x<1ε-1+\varepsilon < x < 1 - \varepsilon, just below the cut,
  • γ4\gamma_4 is the (almost) circular arc z=1+εeiθz = 1 + \varepsilon e^{i\theta} for π<θ<π-\pi < \theta < \pi.

Then we deduce that limε0+Dεz21z4+z2+1dz  =  2i111x2x4+x2+1dx  =  4i011x2x4+x2+1dx \lim_{\varepsilon \to 0+} \int_{D_\varepsilon} \frac{\sqrt{z^2 - 1}}{z^4+z^2+1}\,dz \; = \; -2i\int_{-1}^1 \frac{\sqrt{1-x^2}}{x^4+x^2+1}\,dx \; = \; -4i\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx If we let ΓR\Gamma_R be the circular contour z=Reiθz = Re^{i\theta} for 0θ<2π0 \le \theta < 2\pi, then we have (ΓRDε)z21z4+z2+1dz  =  2πiu{±ω,±ω2}Resz=uz21z4+z2+1 \left(\int_{\Gamma_R} - \int_{D_\varepsilon}\right) \frac{\sqrt{z^2-1}}{z^4 + z^2 + 1}\,dz \; = \; 2\pi i \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{\sqrt{z^2-1}}{z^4+z^2+1} for sufficiently small ε>0\varepsilon>0 and any R>1R > 1. Letting RR \to \infty and ε0+\varepsilon \to 0+, we deduce that 011x2x4+x2+1dx=12πu{±ω,±ω2}Resz=uz21z4+z2+1  =  12πu{±ω,±ω2}Resz=u(z21)z21z61=112πu{±ω,±ω2}u(u21)u21\begin{aligned} \int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx & = \tfrac12\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{\sqrt{z^2-1}}{z^4+z^2+1} \; = \; \tfrac12\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{(z^2-1)\sqrt{z^2-1}}{z^6-1} \\ & = \tfrac{1}{12}\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} u(u^2-1)\sqrt{u^2-1} \end{aligned} If u=eiθu = e^{i\theta} for θ<π|\theta| < \pi, then u+1=2cos12θe12iθu1=2isin12θe12iθu(u21)=2isinθe2iθu+1=2cos12θe14iθu1=2sin12θe14i(θ+πsgn(θ))u21=2sinθe14i(2θ+πsgn(θ))u(u21)u21=2isinθ2sinθe14i(10θ+πsgn(θ))\begin{aligned} u + 1 & = 2\cos\tfrac12\theta e^{\frac12i\theta} \\ u - 1 & = 2i\sin\tfrac12\theta e^{\frac12i\theta} \\ u(u^2-1) & = 2i\sin\theta e^{2i\theta} \\ \sqrt{u+1} & = \sqrt{2\cos\tfrac12\theta} e^{\frac14i\theta} \\ \sqrt{u-1} & = \sqrt{2|\sin\tfrac12\theta|}e^{\frac14i(\theta + \pi\mathrm{sgn}(\theta))} \\ \sqrt{u^2-1} & = \sqrt{2|\sin\theta|} e^{\frac14i(2\theta + \pi\mathrm{sgn}(\theta))} \\ u(u^2-1)\sqrt{u^2-1} & = 2i\sin\theta\sqrt{2|\sin\theta|} e^{\frac14i(10\theta + \pi\mathrm{sgn}(\theta))} \end{aligned} and hence 011x2x4+x2+1dx  =  112π×4×334sin112π  =  π314sin112π  =  π(31)232314 \int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx \; = \; \tfrac{1}{12}\pi \times 4 \times 3^{\frac34} \sin\tfrac{1}{12}\pi \; = \; \frac{\pi}{3^{\frac14}}\sin\tfrac{1}{12}\pi \; = \; \frac{\pi(\sqrt{3}-1)}{2^{\frac32} 3^{\frac14}}

Mark Hennings - 4 years, 5 months ago

Problem 47:

Evaluate,

01arctanx1x2dx\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx

FDP DPF - 4 years, 5 months ago

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The sequence of substitutions u=x2u = x^2, v=1uv = 1-u, w=vw = \sqrt{v}, y=sinhpy = \sinh p and q=epq = e^p give I  =  01tan1x1x2dx=01dx1x201xdy1+x2y2=01dy01xdx1x2(1+x2y2)  =  1201dy01du1u(1+y2u)=1201dy01dvv(1+y2y2v)  =  01dy01dw1+y2y2w2=1201ln(1+y2+y1+y2y)dyy1+y2=0sinh11psinhpdp  =  20sinh11pepdpe2p1=211+2lnqq21dq  =  [lnqln(q+1)+Li2(q)+Li2(1q)]11+2=ln(1+2)ln(2+2)Li2(12)Li2(2)112π2\begin{aligned} I \; = \; \int_0^1 \frac{\tan^{-1}x}{\sqrt{1-x^2}}\,dx & = \int_0^1 \frac{dx}{\sqrt{1-x^2}}\int_0^1 \frac{x\,dy}{1 + x^2y^2} \\ & = \int_0^1\,dy \int_0^1 \frac{x\,dx}{\sqrt{1-x^2}(1 + x^2y^2)} \; = \; \tfrac12\int_0^1\,dy \int_0^1 \frac{du}{\sqrt{1-u}(1+y^2u)} \\ & = \tfrac12\int_0^1\,dy \int_0^1 \frac{dv}{\sqrt{v}(1 + y^2 - y^2v)} \; = \; \int_0^1\,dy \int_0^1 \frac{dw}{1 + y^2 - y^2w^2} \\ & = \tfrac12\int_0^1 \ln\left(\frac{\sqrt{1+y^2} + y}{\sqrt{1+y^2} - y}\right) \frac{dy}{y\sqrt{1+y^2}} \\ & = \int_0^{\sinh^{-1}1} \frac{p}{\sinh p}\,dp \; = \; 2\int_0^{\sinh^{-1}1}\frac{pe^p\,dp}{e^{2p}-1} \\ & = 2\int_1^{1+\sqrt{2}}\frac{\ln q}{q^2-1}\,dq \; = \; -\Big[\ln q \ln(q+1) + \mathrm{Li}_2(-q) + \mathrm{Li}_2(1-q)\Big]_1^{1+\sqrt{2}} \\ & = -\ln(1+\sqrt{2})\ln(2+\sqrt{2}) - \mathrm{Li}_2(-1-\sqrt{2}) - \mathrm{Li}_2(-\sqrt{2}) - \tfrac1{12}\pi^2 \end{aligned} We now use some dilogarithm identities Li2(2)+Li2(12)=13π212ln22iπln2  =  13π218ln2212πiln2Li2(2)=12Li2(2)Li2(2)=12(14π2πiln2)13π2+18ln22+12πiln2+Li2(12)=524π2+18ln22+Li2(12)Li2(12)+Li2(12)=12ln2(2+2)\begin{aligned} \mathrm{Li}_2(\sqrt{2}) + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) & = \tfrac13\pi^2 - \tfrac12\ln^2\sqrt{2} - i\pi\ln\sqrt{2} \; = \; \tfrac13\pi^2 - \tfrac18\ln^22 - \tfrac12\pi i \ln 2 \\ \mathrm{Li}_2(-\sqrt{2}) & = \tfrac12\mathrm{Li}_2(2) - \mathrm{Li}_2(\sqrt{2}) \\ &= \tfrac12\left(\tfrac14\pi^2 - \pi i \ln2\right) - \tfrac13\pi^2 + \tfrac18\ln^22 + \tfrac12\pi i \ln 2 + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) \\ & = -\tfrac{5}{24}\pi^2 + \tfrac18\ln^22 + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) \\ \mathrm{Li}_2(-1-\sqrt{2}) + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) & = -\tfrac12\ln^2(2+\sqrt{2}) \end{aligned} to deduce that Li2(12)+Li2(2)  =  524π2+18ln2212ln2(2+2) \mathrm{Li}_2(-1-\sqrt{2}) + \mathrm{Li}_2(-\sqrt{2}) \; = \; -\tfrac{5}{24}\pi^2 + \tfrac18\ln^22 - \tfrac12\ln^2(2+ \sqrt{2}) and hence that I=ln(1+2)ln(2+2)+524π218ln22+12ln2(2+2)112π2=18π212ln2(1+2)\begin{aligned} I & = -\ln(1 + \sqrt{2})\ln(2 + \sqrt{2}) + \tfrac{5}{24}\pi^2 - \tfrac18\ln^22 + \tfrac12\ln^2(2+\sqrt{2}) - \tfrac{1}{12}\pi^2 \\ & = \tfrac18\pi^2 - \tfrac12\ln^2(1+ \sqrt{2}) \end{aligned}

Mark Hennings - 4 years, 5 months ago

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J=01arctanx1x2dx\displaystyle J=\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx

Perform the change of variable y=1x1+xy=\sqrt{\dfrac{1-x}{1+x}},

J=201arctan(1x21+x2)1+x2dx=201arctan(1)1+x2dx201arctan(x2)1+x2dx=π28201arctan(x2)1+x2dx\begin{aligned}\displaystyle J&=2\int_0^1 \dfrac{\arctan\left(\tfrac{1-x^2}{1+x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\arctan(1)}{1+x^2}dx-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ &=\dfrac{\pi^2}{8}-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ \end{aligned}

01arctan(x2)1+x2dx=[arctanxarctan(x2)]01012xarctanx1+x4dx=π216012xarctanx1+x4dx\begin{aligned} \displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ &=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \end{aligned}

Since,

arctanx=01x1+t2x2dx\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dx

then,

K=012xarctanx1+x4dx=01012x2(1+t2x2)(1+x4)dtdx=0101(2t2(1+t4)(1+x4)+2x2(1+x4)(1+t4)2t2(1+t4)(1+t2x2)dtdx=4(01t21+t4dt)(0111+x4dx)K\begin{aligned} \displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\ \displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\ &=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K \end{aligned}

Therefore,

K=2(01x21+x4dx)(0111+x4dx)\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)

Since,

01x21+x4dx=[142ln(x22x+1x2+2x+1)+122arctan(2x+1)+122arctan(2x1)]01=142(π+ln(322))\begin{aligned}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big) \end{aligned}

and,

0111+x4dx=[142ln(x2+2x+1x22x+1)+122arctan(2x+1)+122arctan(2x1)]01=142(πln(322))\begin{aligned}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big) \end{aligned}

Therefore,

K=π216116(ln(322))2\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}

Therefore,

01arctan(x2)1+x2dx=116(ln(322))2\boxed{\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx=\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}

Therefore,

J=π2818(ln(322))2\boxed{\displaystyle J=\dfrac{\pi^2}{8}-\dfrac{1}{8}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}

FDP DPF - 4 years, 5 months ago

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@Fdp Dpf You do like pulling substitutions out of nowhere! Where did y=1x1+xy = \sqrt{\tfrac{1-x}{1+x}} come from?

This is a nice alternative derivation. Of course, since 322=(1+2)23 - 2\sqrt{2} = (1 + \sqrt{2})^{-2}, our results are the same!

Mark Hennings - 4 years, 5 months ago

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@Mark Hennings I have found out this solution after you write down yours. I was searching but i was stuck. When i have seen your solution i was pretty sure i had missed something about the use of double integration and voilà !.

The inspiration for this problem comes from: http://math.stackexchange.com/questions/2092967/help-to-prove-that-int-0-pi-over-4-arctan-cot2x-mathrm-dx-2-pi2 (the solution i have planned to post initially is the one i have mentioned in a comment)

Your solution is nice as well.

FDP DPF - 4 years, 5 months ago

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@Fdp Dpf To respond your question. When i have an integral 01\int_0^1 i always try the change of variable y=1x1+xy=\dfrac{1-x}{1+x}. Sometimes it's magic. If you do that, in the present integral, it's obvious you have to continue with the change of variable y=xy=\sqrt{x}. My tool to do such computations is Maxima.

FDP DPF - 4 years, 5 months ago

@Mark Hennings That sub is the tangent half angle substitution in algebraic form.

Ishan Singh - 4 years, 5 months ago

* PROBLEM 44 *:

Evaluate 01ln(1+ax1ax)dxx1x21<a<1  . \int_0^1 \ln\left(\frac{1 + ax}{1 - ax}\right) \frac{dx}{x\sqrt{1-x^2}} \hspace{1cm} -1 < a < 1 \;.

Mark Hennings - 4 years, 5 months ago

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For 1<ax<1-1<ax<1, log(1+ax1ax)=2n=0a2n+1x2n+1(2n+1)\log\left(\dfrac{1+ax}{1-ax}\right)=2\sum_{n=0}^{\infty} \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}

01log(1ax1+ax)x1x2dx=201(n=0a2n+1x2n+1(2n+1)1x1x2)dx=2n=0(01a2n+1x2n+1(2n+1)1x1x2dx)=2n=0(a2n+12n+101x2n1x2dx)=n=012n+1B(12,n+12)a2n+1=n=012n+1Γ(12)Γ(n+12)Γ(n+1)a2n+1=n=012n+1(2n)!22nn!πn!a2n+1=πn=0(2nn)22n(2n+1)a2n+1=πarcsin(a)\begin{aligned} \int_0^1 \dfrac{\log\left(\tfrac{1-ax}{1+ax}\right)}{x\sqrt{1-x^2}}dx&=2\int_0^1 \left(\sum_{n=0}^{\infty} \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}\cdot\dfrac{1}{{x\sqrt{1-x^2}}}\right)dx\\ &=2\sum_{n=0}^{\infty}\left(\int_0^1 \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}\cdot\dfrac{1}{{x\sqrt{1-x^2}}} dx\right)\\ &=2\sum_{n=0}^{\infty}\left(\dfrac{a^{2n+1}}{2n+1} \int_0^1 \dfrac{x^{2n}}{\sqrt{1-x^2}}dx\right)\\ &=\sum_{n=0}^{\infty}\dfrac{1}{2n+1}\cdot\text{B}\left(\tfrac{1}{2},n+\tfrac{1}{2}\right)a^{2n+1}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{2n+1}\dfrac{\Gamma\left(\tfrac{1}{2}\right)\Gamma\left(n+\tfrac{1}{2}\right)}{\Gamma(n+1)}a^{2n+1}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{2n+1}\dfrac{\tfrac{(2n)!}{2^{2n}n!}\pi}{n!}a^{2n+1}\\ &=\pi \sum_{n=0}^{\infty} \dfrac{\binom{2n}{n}}{2^{2n}(2n+1)}a^{2n+1}\\ &=\boxed{\pi\arcsin(a)} \end{aligned}

FDP DPF - 4 years, 5 months ago

For variety, if we define F(a)  =  01ln(1+ax1ax)dxx1x21<a<1 F(a) \; = \; \int_0^1 \ln\left(\frac{1 + ax}{1-ax}\right) \frac{dx}{x\sqrt{1-x^2}} \hspace{2cm} -1 < a < 1 then F(a)=01(11+ax+11ax)dx1x2  =  012π(11+acosθ+11acosθ)dθ=0πdθ1+acosθ  =  1202πdθ1+acosθ=12z=122+a(z+z1)dziz  =  iz=1dzaz2+2z+a=iaz=1dz(az+1+1a2)(az+11a2)=2πaResz=a1(1a21)1(az+1+1a2)(az+11a2)  =  π1a2\begin{aligned} F'(a) & = \int_0^1 \left(\frac{1}{1+ax} + \frac{1}{1-ax}\right) \frac{dx}{\sqrt{1-x^2}} \; =\; \int_0^{\frac12\pi} \left(\frac{1}{1 + a\cos \theta} + \frac{1}{1-a\cos\theta}\right)\,d\theta \\ & = \int_0^\pi \frac{d\theta}{1 + a\cos\theta} \; = \; \tfrac12\int_0^{2\pi} \frac{d\theta}{1 + a\cos\theta} \\ & = \tfrac12\int_{|z|=1} \frac{2}{2 + a(z+z^{-1})}\,\frac{dz}{iz} \; = \; -i\int_{|z|=1}\frac{dz}{az^2 + 2z + a} \\ & = -ia \int_{|z|=1}\frac{dz}{\big(az + 1 + \sqrt{1-a^2}\big)\big(az + 1 - \sqrt{1-a^2}\big)} \\ & = 2\pi a\,\mathrm{Res}_{z = a^{-1}(\sqrt{1-a^2}-1)} \frac{1}{\big(az + 1 + \sqrt{1-a^2}\big)\big(az + 1 - \sqrt{1-a^2}\big)} \; = \; \frac{\pi}{\sqrt{1-a^2}} \end{aligned} and hence F(a)  =  0aπ1b2db  =  πsin1a F(a) \; = \; \int_0^a \frac{\pi}{\sqrt{1-b^2}}\,db \; = \; \pi\sin^{-1}a

Mark Hennings - 4 years, 5 months ago

* PROBLEM 48: *

Prove that 0xμ12(x+r)μ(x+s)μdx  =  π(r+s)12μΓ(μ12)Γ(μ) \int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu (x+s)^\mu}\,dx \; = \; \sqrt{\pi}(\sqrt{r}+\sqrt{s})^{1-2\mu}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)} for all r,s>0r,s > 0 and μ>12\mu > \tfrac12.

Mark Hennings - 4 years, 5 months ago

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Someone else can post the next one...

For any a>0a > 0 and 0<b<c0 < b < c and x<1|x| < 1 we have Ia,b,c(x)=01tb1(1t)cb1(1xt)adt=01tb1(1t)cb1(n=0(a)(n)n!xntn)dt=n=0(a)(n)n!xn01tn+b1(1t)cb1dt  =  n=0(a)(n)n!xnB(n+b,cb)=n=0(a)(n)n!xnΓ(n+b)Γ(cb)Γ(n+c)  =  n=0(a)(n)n!xn(b)(n)Γ(b)Γ(cb)(c)(n)Γ(c)=B(b,cb)n=0(a)(n)(b)(n)(c)(n)xnn!  =  B(b,cb)(2F1)(a,b;c;x)\begin{aligned} I_{a,b,c}(x) & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - xt)^{-a}\,dt \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1} \left(\sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n t^n\right)\,dt \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \int_0^1 t^{n+b-1}(1-t)^{c-b-1}\,dt \; = \; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n B(n+b,c-b) \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{\Gamma(n+b) \Gamma(c-b)}{\Gamma(n+c)} \; =\; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{(b)^{(n)} \Gamma(b)\Gamma(c-b)}{(c)^{(n)} \Gamma(c)} \\ & = B(b,c-b)\sum_{n=0}^\infty \frac{(a)^{(n)} (b)^{(n)}}{(c)^{(n)}} \frac{x^n}{n!} \; = \; B(b,c-b) \big({}_2F_1\big)(a,b;c;x) \end{aligned} With the substitution y=t1ty = \tfrac{t}{1-t} we see that 0yb1(1+y)ac(1+αy)ady=01(t1t)b1(11t)ac(1(1α)t1t)adt(1t)2=01tb1(1t)cb1(1(1α)t)adt  =  Ia,b,c(1α)=B(b,cb)(2F1)(a,b;c;1α)\begin{aligned} \int_0^\infty y^{b-1} (1+y)^{a-c} (1 + \alpha y)^{-a}\,dy & = \int_0^1 \left(\frac{t}{1-t}\right)^{b-1}\left(\frac{1}{1-t}\right)^{a-c} \left(\frac{1 - (1-\alpha)t}{1-t}\right)^{-a}\, \frac{dt}{(1-t)^2} \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - (1-\alpha)t)^{-a}\,dt \; = \; I_{a,b,c}(1-\alpha) \\ & = B(b,c-b) \big({}_2F_1\big)(a,b;c;1-\alpha) \end{aligned} for any a>0a>0, 0<b<c0 < b < c and 0<α10 < \alpha \le 1.

Since the integral is symmetric in rr and ss, we may assume that 0<rs0 < r \le s. Then, for any 0<rs0 < r \le s and μ>12\mu > \tfrac12 we have Jμ,r,s=0xμ12(x+r)μ(x+s)μdx  =  rsμ0tμ12(1+t)μ(1+rst)μdt=rsμB(μ+12,μ12)(2F1)(μ,μ+12;2μ;1rs)\begin{aligned} J_{\mu,r,s} & = \int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu(x+s)^\mu}\,dx \; =\; \frac{\sqrt{r}}{s^\mu} \int_0^\infty t^{\mu-\frac12}(1+t)^{-\mu}\big(1 + \tfrac{r}{s}t\big)^{-\mu}\,dt \\ & = \frac{\sqrt{r}}{s^\mu} B\big(\mu+\tfrac12,\mu-\tfrac12\big) \big({}_2F_1\big)\big(\mu,\mu+\tfrac12;2\mu;1-\tfrac{r}{s}\big) \end{aligned} Since (2F1)(a,b;c;z)  =  (1z)aΓ(c)Γ(ba)Γ(b)Γ(ca)(2F1)(a,cb;ab+1;(1z)1)+(1z)bΓ(c)Γ(ab)Γ(a)Γ(cb)(2F1)(b,ca;ba+1;(1z)1) \big({}_2F_1\big)(a,b;c;z) \; = \; \begin{array}{l} \displaystyle(1-z)^{-a} \frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\big({}_2F_1\big)\big(a,c-b;a-b+1;(1-z)^{-1}\big) \\ \displaystyle{}+ (1-z)^{-b} \frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}\big({}_2F_1\big)\big(b,c-a;b-a+1;(1-z)^{-1}\big)\end{array} we deduce that Jμ,r,s=rsμB(μ+12,μ12)[(rs)μΓ(2μ)Γ(12)Γ(μ+12)Γ(μ)(2F1)(μ,μ12;12;sr)+(rs)μ12Γ(2μ)Γ(12)Γ(μ)Γ(μ12)(2F1)(μ+12,μ;32;sr)]=22μ1rμB(μ+12,μ12)[r(2F1)(μ,μ12;12;sr)s(2μ1)(2F1)(μ+12,μ;32;sr)]=πrμΓ(μ12)Γ(μ)[r(2F1)(μ,μ12;12;sr)s(2μ1)(2F1)(μ+12,μ;32;sr)]\begin{aligned} J_{\mu,r,s} & = \frac{\sqrt{r}}{s^\mu}B\big(\mu+\tfrac12,\mu-\tfrac12\big)\left[ \begin{array}{l} \displaystyle\big(\tfrac{r}{s}\big)^{-\mu} \frac{\Gamma(2\mu)\Gamma(\tfrac12)}{\Gamma(\mu+\frac12)\Gamma(\mu)}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) \\ \displaystyle{} + \big(\tfrac{r}{s}\big)^{-\mu-\frac12} \frac{\Gamma(2\mu)\Gamma(-\tfrac12)}{\Gamma(\mu)\Gamma(\mu-\frac12)}\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \end{array} \right] \\ & = \frac{2^{2\mu-1}}{r^\mu}B\big(\mu+\tfrac12,\mu-\tfrac12\big)\left[ \sqrt{r}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) - \sqrt{s}(2\mu-1)\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \right]\\ & = \frac{\sqrt{\pi}}{r^\mu} \frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}\left[ \sqrt{r}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) - \sqrt{s}(2\mu-1)\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \right] \end{aligned} Now standard hypergeometric identities give (2F1)(μ,μ12;12;sr)=12[(1sr)12μ+(1+sr)12μ](2F1)(μ,μ+12;32;sr)=12(2μ1)rs[(1sr)12μ(1+sr)12μ]\begin{aligned} \big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) & = \tfrac12\left[ \left(1 - \sqrt{\tfrac{s}{r}}\right)^{1-2\mu} + \left(1 + \sqrt{\tfrac{s}{r}}\right)^{1-2\mu}\right] \\ \big({}_2F_1\big)\big(\mu,\mu+\tfrac12;\tfrac32;\tfrac{s}{r}\big) & = \frac{1}{2(2\mu-1)}\sqrt{\tfrac{r}{s}}\left[\left(1 - \sqrt{\tfrac{s}{r}}\right)^{1-2\mu} - \left(1 + \sqrt{\tfrac{s}{r}}\right)^{1-2\mu}\right] \end{aligned} and so Jμ,r,s  =  πrμ12Γ(μ12)Γ(μ)(1+sr)12μ  =  πΓ(μ12)Γ(μ)(r+s)12μ J_{\mu,r,s} \; = \; \frac{\sqrt{\pi}}{r^{\mu-\frac12}} \frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}\big(1 + \sqrt{\tfrac{s}{r}}\big)^{1-2\mu} \; = \; \sqrt{\pi}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}(\sqrt{r}+\sqrt{s})^{1-2\mu} as required.

Mark Hennings - 4 years, 5 months ago

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Nice. I'm working on an alternative solution. I hope to get something soon.

FDP DPF - 4 years, 5 months ago

For those having a shot at this one, I need to point out a small, but vital, correction that has been made to the limits of the integral!

Mark Hennings - 4 years, 5 months ago

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It makes things easier :)

FDP DPF - 4 years, 5 months ago

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@Fdp Dpf True :)

Mark Hennings - 4 years, 5 months ago

Please don't post the answer now, post the solution at the end of the week end. Thanks.

FDP DPF - 4 years, 5 months ago

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I was aiming to allow some extra time to compensate for the typo, but will post a solution tomorrow...

Mark Hennings - 4 years, 5 months ago

Problem 49:

Evaluate,

0π6xsin(2x)4sin4x2sin2x+1dx\int_{0}^{\frac{\pi }{6}}\frac{x \mathrm{sin}\left( 2 x\right) }{4 {{\mathrm{sin}^{4}x }}-2 {{\mathrm{sin^{2}}x }}+1}dx

FDP DPF - 4 years, 5 months ago

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For any a>0a > 0 define A(a)=01dxa4+x4  =  1a30a1dz1+z4=12a32[tan1(1+2a1)tan1(12a1)]+14a32ln(a+2+a1a2+a1)B(a)=01x2dxa4+x4  =  1a0a1z2dz1+z4=12a2[tan1(1+2a1)tan1(12a1)]14a2ln(a+2+a1a2+a1)a4A(a)B(a)=18[tan1(1+2a1)tan1(12a1)]2132ln2(a+2+a1a2+a1)\begin{aligned} A(a) & = \int_0^1 \frac{dx}{a^4 + x^4} \; = \; \frac{1}{a^3}\int_0^{a^{-1}} \frac{dz}{1 + z^4} \\ & = \frac{1}{2a^3\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] + \frac{1}{4a^3\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ B(a) & = \int_0^1 \frac{x^2\,dx}{a^4 + x^4} \; = \; \frac{1}{a}\int_0^{a^{-1}} \frac{z^2\,dz}{1 + z^4} \\ & = \frac{1}{2a\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] - \frac{1}{4a\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ a^4A(a)B(a) & = \tfrac{1}{8}\left[ \tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \right]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{aligned} Thus we deduce that (putting z=axz = ax and playing symmetry games) I(a)=0a1xtan1(xa)1+x4dx  =  0a1xdx1+x401xady1+a2x2y2=a40101z2(a4+z4)(a4+y2z2)dydz=12a40101{z2(a4+z4)(a4+y2z2)+y2(a4+y4)(a4+y2z2)}dydz=12a40101y2+z2(a4+y4)(a4+z4)dydz  =  a4A(a)B(a)=18[tan1(1+2a1)tan1(12a1)]2132ln2(a+2+a1a2+a1)\begin{aligned} I(a) & = \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \; = \; \int_0^{a^{-1}} \frac{x\,dx}{1 + x^4} \int_0^1 \frac{\frac{x}{a}\,dy}{1 + a^{-2}x^2y^2} \\ & = a^4 \int_0^1 \int_0^1 \frac{z^2}{(a^4 + z^4)(a^4 + y^2 z^2)}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \left\{ \frac{z^2}{(a^4 + z^4)(a^4 + y^2z^2)} + \frac{y^2}{(a^4 + y^4)(a^4 + y^2z^2)}\right\}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \frac{y^2 + z^2}{(a^4 + y^4)(a^4 + z^4)}\,dy\,dz \; = \; a^4 A(a)B(a) \\ & = \tfrac{1}{8}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{aligned} Moreover J(a)=0a1tan1x2x2+a2dx  =  [1atan1x2tan1(xa)]0a12a0a1xtan1(xa)1+x4dx=1a(tan1a2)22aI(a)\begin{aligned} J(a) & = \int_0^{a^{-1}} \frac{\tan^{-1}x^2}{x^2 + a^2}\,dx \; = \; \Big[\frac{1}{a} \tan^{-1}x^2\,\tan^{-1}\big(\tfrac{x}{a}\big)\Big]_0^{a^{-1}} - \frac{2}{a} \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \\ & = \frac{1}{a}\big(\tan^{-1}a^{-2}\big)^2 - \frac{2}{a}I(a) \end{aligned} Thus, putting u=cos2xu = \cos2x and v=2u13v= \frac{2u-1}{\sqrt{3}}, we see that K=016πxsin2x4sin4x2sin2x+1dx  =  016πxsin2xcos22xcos2x+1dx=14121cos1uduu2u+1  =  14[23tan1(2u13)cos1u]121+1412123tan1(2u13)du1u2=123121tan1(2u13)du1u2  =  12×314013tan1vdv(13v)(3+v)\begin{aligned} K & = \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{4\sin^4x - 2\sin^2x + 1}\,dx \; = \; \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{\cos^22x - \cos2x + 1}\,dx \\ & = \tfrac14\int_{\frac12}^1 \frac{\cos^{-1}u\,du}{u^2 - u + 1} \; = \; \tfrac14\Big[\tfrac{2}{\sqrt{3}}\tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \cos^{-1}u\Big]_{\frac12}^1 + \tfrac14\int_{\frac12}^1 \tfrac{2}{\sqrt{3}} \tan^{-1}\big(\tfrac{2u-1}{\sqrt{3}}\big) \frac{du}{\sqrt{1-u^2}} \\ & = \tfrac{1}{2\sqrt{3}} \int_{\frac12}^1 \tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \frac{du}{\sqrt{1-u^2}} \; = \; \frac{1}{2 \times 3^{\frac14}} \int_0^{\frac{1}{\sqrt{3}}} \frac{\tan^{-1}v\,dv}{\sqrt{(1 - \sqrt{3}v)(\sqrt{3}+v)}} \end{aligned} The key substitution w  =  13v3+v w \; = \; \sqrt{\frac{1 - \sqrt{3}v}{\sqrt{3}+v}} yields K=13140314tan1(13w23+w2)dw3+w2  =  13140314[16πtan1w2]dw3+w2=π6×314[314tan1(w314)]0314314J(314)=π2363314[314(tan1312)22×314I(314)]  =  23I(314)=143[tan1(2314314)]21163ln2(314+2+3143142+314)\begin{aligned} K & = \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \tan^{-1}\left(\frac{1 - \sqrt{3}w^2}{\sqrt{3} + w^2}\right) \frac{dw}{\sqrt{3} + w^2} \; = \; \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \Big[\tfrac16\pi - \tan^{-1}w^2\Big] \frac{dw}{\sqrt{3} + w^2} \\ & = \frac{\pi}{6 \times 3^{\frac14}} \Big[3^{-\frac14} \tan^{-1}\big(\tfrac{w}{3^{\frac14}}\big)\Big]_0^{3^{-\frac14}} - 3^{-\frac14}J(3^{\frac14}) \\ & = \frac{\pi^2}{36\sqrt{3}} - 3^{-\frac14}\Big[3^{-\frac14}\left(\tan^{-1}3^{-\frac12}\right)^2 - 2 \times 3^{-\frac14} I\big(3^{\frac14}\big)\Big] \; = \; \tfrac{2}{\sqrt{3}} I\big(3^{\frac14}\big) \\ & = \frac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right) \end{aligned} noting that tan1(1+2a1)tan1(12a1)  =  tan1(2aa1)a>1 \tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \; =\; \tan^{-1}\left(\frac{\sqrt{2}}{a - a^{-1}}\right) \hspace{1cm} a > 1

Mark Hennings - 4 years, 5 months ago

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My solution is close to yours. I will post it later (01 AM here) . This integral is based on the solution i have provided for problem 47. I have played around with it and i have build this one.

FDP DPF - 4 years, 5 months ago

Actually my solution is the same than yours. The only difference is the way you manage to transform the trigonometric integral.

FDP DPF - 4 years, 4 months ago

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@Fdp Dpf Yes, but your method of transforming the original integral into what is, apart from a variable scaling, 23I(31/4)\tfrac{2}{\sqrt{3}} I(3^{1/4}) (in my notation), is elegant.

Mark Hennings - 4 years, 4 months ago

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@Mark Hennings if F(a,t,x)=x(tx+a)(x2+a)F(a,t,x)=\frac{x}{\left( t\cdot x+a\right) \cdot \left( {{x}^{2}}+a\right) }, compute 01F(1,t,x)dt\int_0^1 F(1,t,x)dt isn't nice?

Problem 47 relies on the evaluation of 0101F(1,t2,x2)dtdx\int_0^1 \int_0^1 F(1,t^2,x^2) dtdx and problem 49 relies on 0101F(3,t2,x2)dtdx\int_0^1 \int_0^1 F(3,t^2,x^2) dtdx

Who will post problem 50, the last one?

FDP DPF - 4 years, 4 months ago

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@Fdp Dpf I'll post by evening, in 5 hours.

Ishan Singh - 4 years, 4 months ago

@Fdp Dpf Let me clarify. Both of our methods evaluate 23I(314)\tfrac{2}{\sqrt{3}}I(3^{\frac14}). What I was saying was that your technique for converting the original integral to 23I(314)\tfrac{2}{\sqrt{3}}I(3^{\frac14}) was more elegant than mine - in Mathematics, the word "elegant" is complimentary!

Mark Hennings - 4 years, 4 months ago

Aliter,

Let.

I=0π6xsin(2x)4sin4x2sin2x+1 dx\text{I} = \int_{0}^{\frac{\pi}{6}} \dfrac{x \sin(2x)}{4\sin^4 x - 2\sin^2 x + 1} \ \mathrm{d}x

=14(αβ)0π6xsin(2x)(1cos(2x)1+β1cos(2x)1+α) = \dfrac{1}{4(\alpha - \beta)} \int_{0}^{\frac{\pi}{6}} x \sin (2x) \left(\dfrac{1}{\cos(2x) - 1 + \beta} - \dfrac{1}{\cos(2x) - 1 + \alpha}\right)

where α\alpha and β\beta are the roots of the equation t212t+14=0 t^2 - \dfrac{1}{2} t + \dfrac{1}{4} = 0

Now, from Problem 25, we have,

sinxp2+2pcosx+1=k=1(p)k1sin(kx) \dfrac{\sin x}{p^2 +2p \cos x +1} = \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)

Using this and interchanging, sum and integral ( and after a lot of straight forward simplification/calculations involving dilogarithm and logarithm) we get,

I=143[tan1(2314314)]21163ln2(314+2+3143142+314)\text{I} = \dfrac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right)

Ishan Singh - 4 years, 5 months ago

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Just pipped you! You might want to clarify what α\alpha and β\beta are, though.

Why don't you post the last integral!

Mark Hennings - 4 years, 5 months ago

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@Mark Hennings Thanks for letting me do the honors. I'll post it tomorrow, I'll need some time to create a nice one :)

Ishan Singh - 4 years, 5 months ago

What was your method?

Ishan Singh - 4 years, 5 months ago

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I am interested. My solution used/generalised a lot of the tricks that FDP uses in his solutions, so it will be interesting if he has a different approach.

Mark Hennings - 4 years, 5 months ago

J=0π6xsin(2x)12sin2x+4sin4xdx=0π62xsinxcosx(1sin2x)2+3sin4xdx=0π62xsinxcosxcos4+3sin4xdx=0π62xtanxcos2x(1+3tan4x)dx\begin{aligned}J&=\int_0^{\tfrac{\pi}{6}} \dfrac{x\sin(2x)}{1-2\sin^2 x+4\sin^4 x}dx\\ &= \int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{(1-\sin^2 x)^2+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{\cos^4+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\tan x}{\cos^2 x(1+3\tan^4 x)}dx\\ \end{aligned}

Perform the change of variable y=3tanxy=\sqrt{3}\tan x,

J=012xarctan(x3)3+x4dxJ=\int_0^1 \dfrac{2x\arctan\left(\tfrac{x}{\sqrt{3}}\right)}{3+x^4}dx

Since,

arctanu=01u1+t2u2dt\arctan u=\int_0^1 \dfrac{u}{1+t^2u^2}dt

J=230101x2(3+x4)(3+t2x2)dtdx=230101(t2+x2(t4+3)(x4+3)t2(t4+3)(t2x2+3))dtdx=430101t2(3+t4)(3+x4)dtdxJ\begin{aligned}J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{x^2}{(3+x^4)(3+t^2x^2)}dtdx\\ &=2\sqrt{3}\int_0^1 \int_0^1 \left(\frac{{{t}^{2}}+{{x}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{x}^{4}}+3\right) }-\frac{{{t}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{t}^{2}}{{x}^{2}}+3\right) }\right)dtdx\\ &=4\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx-J\\ \end{aligned}

therefore,

J=230101t2(3+t4)(3+x4)dtdx=23(01x23+x4dx)(0113+x4dx)\begin{aligned} J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx\\ &=2\sqrt{3}\left(\int_0^1 \dfrac{x^2}{3+x^4}dx \right)\left(\int_0^1 \dfrac{1}{3+x^4}dx\right)\\ \end{aligned}

Since,

0113+x4dx=[arctan(2314x3x2)232314log(32314x+x2x2+2314x+3)252314]01=1232334arctan(22314+22334)1252334log(1+323141+3+2314)\begin{aligned} \int_0^1 \dfrac{1}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}{{3}^{\frac{1}{4}}}}-\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot{{3}^{\frac{1}{4}}} x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}{{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(\dfrac{1+\sqrt{3}-\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{1+\sqrt{3}+\sqrt{2}\cdot 3^{\tfrac{1}{4}}}\right) \end{aligned}

1+3+33141+\sqrt{3}+\sqrt{3}\cdot 3^{\tfrac{1}{4}} is a root of the polynomial X44X316X+16X^4-4X^3-16X+16,

Therefore,

0113+x4dx=1232334arctan(231431)1252334log(1+32231422334)\int_0^1 \dfrac{1}{3+x^4}dx=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{\sqrt{3}-1}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right)

and,

01x23+x4dx=[arctan(2314x3x2)232314+log(32314x+x2x2+2314x+3)252314]01=1232314arctan(22314+22334)+1252314log(1+32231422334)\begin{aligned} \int_0^1 \dfrac{x^2}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{1}{4}}}}+\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}\cdot {{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{1}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)+\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{1}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) \end{aligned}

Therefore,

J=143arctan2(22314+22334)1163log2(1+32231422334)\boxed{J=\dfrac{1}{4\sqrt{3}}\arctan^2\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{16\sqrt{3}}\log^2\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) }

PS:

F(x,t)=x(a+x2)(a+tx)=ta+t21a+x2+1a+t2xa+x2t(a+t2)(a+tx)F(x,t)=\dfrac{x}{(a+x^2)(a+tx)}=\dfrac{t}{a+t^2}\cdot \dfrac{1}{a+x^2} +\dfrac{1}{a+t^2}\cdot \dfrac{x}{a+x^2}-\dfrac{t}{(a+t^2)(a+tx)}

is a nice function.

FDP DPF - 4 years, 4 months ago

Ohh!! I missed this season. I even didn't know that this happened. :(

Surya Prakash - 4 years, 1 month ago

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Maybe this time you can host it :)

Ishan Singh - 4 years, 1 month ago

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When is starting the season 4? :)

FDP DPF - 4 years, 1 month ago

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@Fdp Dpf I think in December 2017.

Ishan Singh - 4 years, 1 month ago

Problem 50 :

Evaluate 0sin(2nx)cos(2arctan(ax))(1+a2x2)sinx dx \int_{0}^{\infty} \dfrac{\sin(2nx) \cos(2\arctan(ax))}{(1+a^2x^2) \sin x} \ \mathrm{d}x

where a>0a>0 , nZ+n \in \mathbb{Z^+}

Ishan Singh - 4 years, 4 months ago

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Suppose that a>0a > 0 and nNn \in \mathbb{N}. Note that I  =  0sin(2nx)cos(2tan1(ax))(1+a2x2)sinxdx  =  01a2x2(1+a2x2)2sin2nxsinxdx  =  limδ0+Iδ I \; = \; \int_0^\infty \frac{\sin(2nx)\cos\big(2\tan^{-1}(ax)\big)}{(1 + a^2x^2)\sin x}\,dx \; = \; \int_0^\infty \frac{1-a^2x^2}{(1 + a^2x^2)^2}\frac{\sin 2nx}{\sin x}\,dx \; = \; \lim_{\delta \to 0+}I_\delta where Iδ  =  01a2(x+δ)2(1+a2(x+δ)2)2sin2nxsinxdx I_\delta \; = \; \int_0^\infty \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\,dx where we can use the Dominated Convergence Theorem to justify the limit, since 1a2(x+δ)2(1+a2(x+δ)2)2sin2nxsinx    11+a2(x+δ)2sin2nxsinx    Nn1+a2x2 \left| \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \left| \frac{1}{1 + a^2(x+\delta)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \frac{N_n}{1 + a^2x^2} for all x0x \ge 0 and δ>0\delta > 0, where sin2nxsinxNn\big|\tfrac{\sin 2nx}{\sin x}\big| \le N_n for all x0x \ge 0. It is trivial that 0tcos(ta)extdt  =  a2a2x21(1+a2x2)2a,x>0 \int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-xt}\,dt \; = \; a^2 \frac{a^2x^2 - 1}{(1 + a^2x^2)^2} \hspace{2cm} a,x > 0 and so Iδ=a20(0tcos(ta)e(x+δ)tdt)sin2nxsinxdx=a20tcos(ta)eδt(0extsin2nxsinxdx)dt\begin{aligned} I_\delta & = -a^{-2}\int_0^\infty\left(\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-(x+\delta)t}\,dt\right) \frac{\sin 2nx}{\sin x}\,dx \\ & = -a^{-2}\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx\right)\,dt \end{aligned} Since 0extsin(2m+2)xsinxdx0extsin2mxsinxdx  =  20extcos(2m+1)xdx  =  2tt2+(2m+1)2 \int_0^\infty e^{-xt} \frac{\sin(2m+2)x}{\sin x}\,dx - \int_0^\infty e^{-xt} \frac{\sin 2mx}{\sin x}\,dx \; = \; 2\int_0^\infty e^{-xt} \cos(2m+1)x\,dx \; = \; \frac{2t}{t^2 + (2m+1)^2} a simple induction shows that 0extsin2nxsinxdx  =  2tk=0n11t2+(2k+1)2t>0 \int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx \; = \; 2t\sum_{k=0}^{n-1} \frac{1}{t^2 + (2k+1)^2} \hspace{2cm} t > 0 and hence Iδ=2a20cos(ta)eδt(k=0n1t2t2+(2k+1)2)dt=2a20cos(ta)eδt(nk=0n1(2k+1)2t2+(2k+1)2)dt=2a2[nδδ2+a2k=0n1(2k+1)20cos(ta)eδtt2+(2k+1)2dt]\begin{aligned} I_\delta & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\sum_{k=0}^{n-1} \frac{t^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left( n - \sum_{k=0}^{n-1} \frac{(2k+1)^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2}\left[ \frac{n\delta}{\delta^2 + a^{-2}} - \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big) e^{-\delta t}}{t^2 + (2k+1)^2}\,dt \right] \end{aligned} and hence, letting δ0+\delta \to 0+, I=2a2k=0n1(2k+1)20cos(ta)t2+(2k+1)2dt=2a2k=0n1(2k+1)2×π2(2k+1)e2k+1a=πa2k=0n1(2k+1)e2k+1a\begin{aligned} I & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big)}{t^2 + (2k+1)^2}\,dt \\ & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \times \frac{\pi}{2(2k+1)}e^{-\frac{2k+1}{a}} \\ & = \frac{\pi}{a^2}\sum_{k=0}^{n-1} (2k+1)e^{-\frac{2k+1}{a}} \end{aligned} using a standard Fourier transform to finish off. This sum can be calculated, of course, but the end result is not particularly pretty, do I shall leave things here.

Mark Hennings - 4 years, 4 months ago

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Nice! I used the expansion sin(2nx)2sinx=k=1ncos((2k1)x) \dfrac{\sin(2nx)}{2\sin x} = \sum_{k=1}^{n} \cos((2k-1)x)

The result can also be generalized to 0sin(2nx)cos(parctan(ax))(1+a2x2)p2sinx dx ; nZ,a,p>0 \int_{0}^{\infty} \dfrac{\sin(2nx) \cos(p\arctan(ax))}{(1+a^2x^2)^{\frac{p}{2}} \sin x} \ \mathrm{d}x \ ; \ n \in \mathbb{Z}, a,p>0

Ishan Singh - 4 years, 4 months ago

Let Cn1(a)C_{n-1}(a) be the previous integral. I suspect one can find a closed form for the ordinary generating function for the sequence (Cn(a))\left(C_n(a)\right) because U2n+1(cosx)=sin((2n+2)xsinxU_{2n+1}(\cos x)=\dfrac{\sin((2n+2)x}{\sin x}, UjU_j being the jj-nth Chebyshev polynomial of the second kind.

FDP DPF - 4 years, 4 months ago

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I looked hard on the web and could not find a closed expression for the Laplace transform of the UnU_n, which would have been really useful!

Mark Hennings - 4 years, 4 months ago
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