Brilliant Integration - Season 3(Part 1)
Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel ofThe 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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Problem 26 :
Show that ∫0∞ln(bs+xsas+xs)dx=π(a−b)cosecsπa,b>0,s>1.
This problem has been solved by Fdp Dpf.
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Solution problem 26:
Let , F(a)=∫0+∞ln(bs+xsas+xs)dx
Observe that, F(b)=0
F′(a)=sas−1∫0+∞as+xs1dx
Perform the change of variable,
y=ax,
F′(a)=s∫0+∞1+xs1dx
It's well known that,
∫0+∞1+xs1dx=sπcosec(sπ)
Therefore,
F′(a)=πcosec(sπ)
Therefore, F(a)=πcosec(sπ)a+k ,
k a real constant.
Since F(b)=0 then k=−πcosec(sπ)b
Therefore,
F(a)=π(a−b)cosec(sπ)
Proof of:
∫0+∞1+xs1dx=sπcosec(sπ)
Perform the change of variable
y=xs
∫0+∞1+xs1dx=s1∫0+∞1+xxs1−1dx
∫0+∞1+xxs1−1dx=β(s1,1−s1)=Γ(1)Γ(s1)Γ(1−s1)=sin(sπ)π=πcosec(sπ)
β being the Beta Euler function. Third line is the use of Euler's reflection formula.
Problem 25 :
Prove That
π1∫0πxarctan(1+pcosxpsinx) dx=Li2(p)∀ ∣p∣<1
Notation : Li2(z) denotes the Dilogarithm Function.
This problem has been solved by Mark Hennings.
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If we define g(p)=∫0πln(1+2pcosx+p2)dx∣p∣<1 then g(0)=0 and, using the complex subsitution z=eix, g′(p)=∫0π1+2pcosx+p22cosx+2pdx=21∫−ππ1+2pcosx+p22cosx+2pdx=21∫∣z∣=11+p(z+z−1)+p2z+z−1+2pizdz=2i1∫∣z∣=1z(pz+1)(z+p)z2+2pz+1dz=π(Resz=0+Resz=−p)z(pz+1)(z+p)z2+2pz+1=π(p1−p(1−p2)1−p2)=0 so that g(p)=0 for all ∣p∣<1. If we now define f(p)=π1∫0πxtan−1(1+pcosxpsinx)dx∣p∣<1 then f(0)=0 and f′(p)=π1∫0π1+(1+pcosxpsinx)2x×(1+pcosx)2sinxdx=π1∫0π1+2pcosx+p2xsinxdx=π1[−2pxln(1+2pcosx+p2)]0π+2πp1∫0πln(1+2pcosx+p2)dx=−pln(1−p)+g(p)=−pln(1−p) and hence it follows that f(p)=Li2(p).
Aliter,
Let f(p)=π1∫0πxarctan(1+pcosxpsinx) dx
Firstly, we have f(0)=0
Note that,
k=1∑∞(−p)k−1sin(kx)=p1ℑ(k=0∑∞(−peix)k)
=p1ℑ(1+peix1)
=p2+2pcosx+1sinx
where ℑ(z) denotes the imaginary part of z.
So we have,
f′(p)=π1∫0πp2+2pcosx+1xsinx dx
=π1∫0πx(k=1∑∞(−p)k−1sin(kx)) dx
=π1k=1∑∞(−p)k−1∫0πxsin(kx) dx
=π1k=1∑∞(−p)k−1(k2sin(kπ)−kπcos(kπ))
Since sin(kπ)=0 and cos(kπ)=(−1)k ∀ k∈Z, we get,
f′(p)=k=1∑∞kpk−1=−pln(1−p)
∴f(p)=Li2(p) □
Problem 27:
Show that
∫06−2+16−2−1(x−1)x2−2(15+83)x+1lnxdx=32(2−3)G
This problem has been solved by Mark Hennings.
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Put α=121π. With the substitution x=1+cosθ1−cosθ, we obtain after much manipulation that I=∫06−2+16−2−1(x−1)x2−2(15+83)x+1)lnxdx=−21sinα∫0αcosθcos2θ−cos2αln(1+cosθ1−cosθ)sinθdθ Putting u=cosθ now gives I=−21sinα∫cosα1uu2−cos2αln(1+u1−u)du Now putting u=cosαsecϕ yields I=−21tanα∫0αln(cosϕ+cosαcosϕ−cosα)dϕ=−21tanα∫0αln(tan(2α+ϕ)tan(2α−ϕ))dϕ=−21tanα∫0α{ln(tan(2α+ϕ))+ln(tan(2α−ϕ))}dϕ=−tanα{∫21ααln(tanϕ)dϕ−∫21α0ln(tanϕ)dϕ}=−tanα∫0αln(tanϕ)dϕ=−tanα∫0121πln(tanϕ)dϕ=32tanαG=32(2−3)G using a standard integral representation of G at the very last stage.
Problem 29:
Show that 0∫∞x(x2+1)e−txdx=π[2C(π2t)sint−2S(π2t)cost−sin(t−π/4)]
where C and S are Fresnel cosine integral and Fresnel sine integral, respectively.
This problem has been solved by Mark Hennings.
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In this question, we are using the Fresnel integrals C(x)=∫0xcos(21πt2)dtS(x)=∫0xsin(21πt2)dt If we define A(t)=cost∫0∞x2+1e−txxdx−sint∫0∞x(x2+1)e−txdx for t>0 then A′(t)=[t]l−sint∫0∞x2+1e−txxdx−cost∫0∞x2+1e−txxxdx−cost∫0∞x(x2+1)e−txdx+sint∫0∞x2+1e−txxdx=−cost∫0∞xe−txdx=−tπcost for any t>0, so that dtd[A(t)+π2C(π2t)]=0 Since A(t)→0 as t→∞, we deduce that A(t)+π2C(π2t)=21πt>0 Similarly, if we define B(t)=sint∫0∞x2+1e−txxdx+cost∫0∞x(x2+1)e−txdx for t>0, then we can show that B′(t)=−sint∫0∞xe−txdx=−tπsint and hence B(t)+π2S(π2t)=21πt>0 Thus we deduce that ∫0∞x(x2+1)e−txdx=B(t)cost−A(t)sint=cost[21π−π2S(π2t)]−sint[21π−π2C(π2t)]=π2C(π2t)−π2S(π2t)−πsin(t−41π) as required.
Problem 30:
Show that ∫01ln(1+41x2+21x)xdx=201π2
This problem has been solved by Fdp Dpf.
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I=∫01xln(1+4x2+2x)dx
Perform the change of variable y=2x,
I=∫021xln(1+x2+x)dx
Perform the change of variable y=arsinh(x),
I=∫0arsinh(21)sinh(x)xcosh(x)dx=∫0arsinh(21)1−e−2xx(1+e−2x)dx=∫0arsinh(21)(x(1+e−2x)n=0∑+∞e−2nx)dx=2(arsinh(21))2+2∫0arsinh(21)(xn=1∑+∞e−2nx)dx=2(arsinh(21))2+2n=1∑+∞(∫0arsinh(21)xe−2nx)dx=2(arsinh(21))2+21n=1∑+∞(n21−n2(1+2arsinh(21)n)e−2arsinh(21)n)=2(arsinh(21))2+21ζ(2)−21Li2(e−2asinh(21))+arsinh(21)ln(1−e−2arsinh(21))
Since, arsinh(21)=ln(21+25) then,
I=12π2−21Li2(23−25)−21(ln(21+25))2
Since , Li2(23−25)=15π2−(ln(21+25))2 then,
I=12π2−30π2=20π2
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Nice. A bit of integration by parts makes the earlier stage a little clearer, perhaps...
Note that dxdln(1+41x2+21x)=1+41x2+21x211+41x221x+21=4+x21 and so, integrating by parts, ∫01ln(1+41x2+21x)xdx=[(lnx)ln(1+41x2+21x)]01−∫014+x2lnxdx=−∫014+x2lnxdx Since sinh−121=ln(21(5+1)), the substitution x=2sinhu yields ∫014+x2lnxdx=∫0sinh−121ln(2sinhu)du=∫0sinh−121[u+ln(1−e−2u)]du=21ln2(25+1)+∫0ln(25+1)e−2uln(1−e−2u)e−2udu=21ln2(25+1)−21∫123−5vln(1−v)dv using the substitution v=e−2u. Thus ∫014+x2lnxdx=21ln2(25+1)+21Li2(23−5)−21Li2(1)=301π2−121π2=−201π2 and so ∫01ln(1+41x2+21x)xdx=201π2
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∫ln(1+41x2+21x)dx=xln(1+41x2+21x)−x2+4
You're right, it's easiest this way. By the way,Problem 31:
Show that
∫01(x+1)lnxx−1dx=ln(2π)
This problem has been solved by Ishan Singh.
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Let
f(p)=∫01(x+1)lnxxp−1 dx;p>0
⟹f′(p)=∫01x+1xpdx
=∫01x2−1xp(x−1)dx
Substitute x2↦x and simplify to get,
f′(p)=21∫01(x−1x2p−1−x−1x2p−1−1)dx
=21(ψ(2p+1)−ψ(2p−1+1))
Note that f(0)=0
⟹f(1)=∫01f′(p) dp
=21∫01(ψ(2p+1)−ψ(2p−1+1))dp
=log(2π)
where the last equality follows since ψ(z)=dzdlog(Γ(z))
Problem 37:
Evaluate ∫021πln(cosx+sinx)dx
This problem has been solved by Fdp Dpf.
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Solution to problem 37.
Since,
sin(x+4π)=cos(4π)sinx+sin(4π)cosx=22(sinx+cosx)
then,
∫02πln(cosx+sinx)dx=∫02πln(2sin(x+4π))dx=∫04πln(2sin(x+4π))dx+∫4π2πln(2sin(x+4π))dx=∫04πln(2sin((4π−x)+4π))dx+∫04πln(2sin((x+4π)+4π))dx=2∫04πln(2cosx)dx=4πln2+2∫04πln(cosx)dx=4πln2−∫04πln((cosx)21)dx=4πln2−∫04πln(1+(tanx)2)dx
In the latter integral perform the change of variable y=tanx, therefore, ∫02πln(cosx+sinx)dx=4πln2−∫011+x2ln(1+x2)dx
∫0+∞1+x2ln(1+x2)dx=∫011+x2ln(1+x2)dx+∫0+∞1+x2ln(1+x2)dx
In the latter integral perform the change of variable y=x1, therefore,
∫0+∞1+x2ln(1+x2)dx=∫011+x2ln(1+x2)dx+∫011+x2ln(1+x21)dx=2∫011+x2ln(1+x2)dx−2∫011+x2lnxdx=2∫011+x2ln(1+x2)dx+2G
G being the Catalan constant. But, ∫0+∞1+x2ln(1+x2)dx=πln2
(see my answer to problem 36)
Therefore, ∫011+x2ln(1+x2)dx=2πln2−G
Therefore,
∫02πln(cosx+sinx)dx=4πln2−(2πln2−G)=G−4πln2
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Aliter,
The integral ∫04πlog(cosx) dx can also be done using the identity log(cosx)=−log2−k=1∑∞kcos(2kx).
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cos(2kx) as 2e2ikx+e−2ikx and then using the Taylor Series of logarithm. Btw, Happy New Year.
I like solutions using elementary methods too. I consider the above identity elementary, it can be proved by writingProblem 28:
Show that,
∫01xarctan(2−xx(1−x))dx=3G
G being the Catalan constant.
This problem has been solved by Mark Hennings.
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Your derivation of this result on MSE is pretty compact, so I shall expand it. Suppose that a>1, and let b=a−11+a2. The substitution y=x+bax gives ∫01tan−1(x+bax)xdx=∫0a+1a−1tan−1ybya−y(a−y)2abdy=∫0a+1a−1y(a−y)atan−1ydy while the substitution z=b−axx gives ∫01tan−1(ax−bx)xdx=−∫0a+1a−1tan−1zbzaz+a(az+1)2bdz=−∫0a+1a−1z(az+1)1tan−1zdz and hence ∫01{tan−1(ax+bax)+tan−1(ax−bx)}xdx=∫0a+1a−1(y(a−y)a−y(ay+1)1)tan−1ydy=∫0a+1a−1(a−y)(ay+1)a2+1tan−1ydy=∫)a+1a−1(a−y1+1+aya)tan−1ydy=[ln(a−y1+ay)tan−1y]0a+1a−1+∫0a+1a−11+y2ln(1+aya−y)dy=∫0a+1a−11+y2ln(1+aya−y)dy=∫1a1+x2lnxdx Now tan[tan−1(x+bax)+tan−1(ax−bx)]=1−x+baxax−bxx+bax+ax−bx=(x+b)(ax−b)−ax2ax(ax−b)+x(x+b)=(a−1)21+a2−xx(1−x) and hence it follows that ∫01tan−1((a−1)21+a2−xx(1−x))xdx=∫1a1+x2lnxdx Putting a=2+3 yields ∫01tan−1(2−xx(1−x))xdx=∫12+31+x2lnxdx=∫41π125πln(tanθ)dθ=−∫121π41πln(tanϕ)dϕ=−((−G)−(−32G))=31G using the final substitution ϕ=21π−θ.
I won't be posting the next problem until after Christmas!
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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 !
Problem 36:
Show that,
∫0+∞1+x4xln(1+2x+x2)dx=2πln2
This problem has been solved by Mark Hennings.
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The substitutions t=1−s and s=sinθ give ∫01(t(2−t)1−1)1−tdt=∫01(1−s21−1)sds=∫021πtan21θdθ=ln2 The substitution x2=2−tt and one of the standard integral representations of the Catalan constant G gives ∫01t(2−t)1(41π−tan−12−tt)1−tdt=2∫01(41π−tan−1x)1−x2dx=G These results together give us that ∫0∞x2+2x+1ln(x2+2x+1)dx=∫0∞x2+2x+1dx∫01tx2+t2x+1x2+2xdt=∫01∫0∞(tx2+t2x+1)(x2+2x+1)x2+2xdxdt=∫011−tdt∫0∞(tx2+t2x+11−x2+2x+11)dx=∫011−tdt{t(2−t)2(21π−tan−12−tt)−22π}=∫011−tdt{t(2−t)2(41π−tan−12−tt)+22π(t(2−t)1−1)} and so ∫0∞x2+2x+1ln(x2+2x+1)dx==2G+22πln2 Standard substitutions show that f(a)=∫0∞(x4+1)ax2+1dx=41[B(43,a−43)+B(41,a−41)] for a>43, and hence f′(a)=4Γ(a)Γ(43)Γ(a−43)ψ(a−43)+Γ(41)Γ(a−41)ψ(a−41)−4Γ(a)Γ(43)Γ(a−43)+Γ(41)Γ(a−41)ψ(a) and hence ∫0∞x4+1(x2+1)ln(x4+1)dx=−f′(1)=−22π[ψ(41)+ψ(43)−2ψ(1)]=23πln2 Next, the substitution x=tan21θ and another standard integral representation of the Catalan constant gives ∫0∞1+x41+x2ln(x2−2x+1x2+2x+1)dx=∫0πln(1−21sinθ1+21sinθ)1+cos2θdθ=2∫021πln(1−21sinθ1+21sinθ)1+cos2θdθ=22G Thus ∫0∞(x2+2x+1ln(x2+2x+1)+x2−2x+1ln(x2+2x+1))dx=2∫0∞x4+1x2+1ln(x2+2x+1)dx=∫0∞x4+1x2+1[ln(x2−2x+1x2+2x+1)+ln(x4+1)]dx=22G+23πln2 and hence ∫0∞x2−2x+1ln(x2+2x+1)dx=(22G+23πln2)−(2G+22πln2)=2G+225πln2 Finally (!) we deduce that ∫0∞x4+1xln(x2+2x+1)dx=221∫0∞(x2−2x+1ln(x2+2x+1)−x2+2x+1ln(x2+2x+1))=21πln2
Alternative solution to problem 36
Let,
I=∫0+∞1+x4x(ln(1+2x+x2)dx
J=∫0+∞1+x4x(ln(1−2x+x2)dx
I+J=∫0+∞1+x4xln(1+x4)dx
Perform the change of variable y=x2,
I+J=21∫0+∞1+x2ln(1+x2)dx
Perform the change of variable y=arctanx,
I+J=−∫02πln(cosx)dx
Perform the change of variable y=2π−x,
∫02πln(cosx)dx=∫02πln(sinx)dx
Therefore,
2(I+J)=−∫02πln(cosx)dx−∫02πln(sinx)dx=−∫02πln(cos(x)sin(x))dx=−∫02πln(2sin(2x))dx=−∫02πln(sin(2x))dx+2πln2
In the latter integral perform the change of variable y=2x,
2(I+J)=−21∫0πln(sinx)dx+2πln2=−21∫02πln(sinx)dx−21∫2ππln(sinx)dx+2πln2
In the latter integral perform the change of variable y=π−x, therefore,
2(I+J)=−21∫02πln(sinx)dx−21∫02πln(sinx)dx+2πln2=−∫02πln(sinx)dx+2πln2=−∫02πln(cosx)dx+2πln2=I+J+2πln2
Therefore,
I+J=2πln2
Define on [0;2],
F(a)=∫0+∞1+x4x(ln(1+ax+x2)−ln(1−ax+x2))dx
Observe that, F(2)=I−J and F(0)=0,
F′(a)=∫0+∞(x2−ax+1)(x2+ax+1)(x4+1)2x(1+x2)dx
F′(a)=⎣⎢⎢⎡4−a2⋅(a2−2)2(arctan(4−a22x−a)+arctan(4−a22x+a))−a2−22(arctan(2x−1)+arctan(2x+1))⎦⎥⎥⎤0+∞
Therefore,
F′(a)=(a2−2)4−a22π−a2−22π
F(2)=∫02F′(a)da=2π[ln((2−a)(4−a2+a)(2+a)(4−a2−a))]02=2πln2
Therefore,
I+J=I−J=2πln2
and,
I=2πln2 J=0
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You could speed up your calculation of I+J by doing a Beta function trick. It is equal to −41dadB(21,a−21)∣∣∣a=1
I came up with a similar solution just now, and was about to post, but you beat me to it by 4 hours!
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Problem 32 : Prove That
∫0∞(1+x21−x2)⋅ln(x)⋅arctan(x2+1x) xdx=−20π3
This problem has been solved by Mark Hennings.
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Suppose that u>0. Integrating by parts, we see note that ∫0∞1+41u2sech2ttsechttanhtdt=[−u2ttan−1(21usecht)]0∞+u2∫0∞tan−1(21usecht)dt=u2∫0∞tan−1(21usecht)dt Now the substitution v=21usecht yields ∫0∞1+41u2sech2ttsechttanhtdt=u2∫21u0vu2−4v2tan−1v(−u)dv=2∫021uvu2−4v2tan−1vdv=u2∫01w1−w2tan−1(21uw)dw=uπln(1+41u2+21u) Note that the identity ∫01w1−w2tan−1pwdw=21πln(1+p2+p) is a standard integral from Gradshteyn & Ryzhik at the last stage. Thus, using the substitution x=et, ∫0∞1+x21−x2lnxtan−1(x2+1x)xdx=−∫Rttanhttan−1(21secht)dt=−2∫0∞ttanhttan−1(21secht)dt=−2∫0∞ttanht(∫011+41u2sech2t21sechtdu)dt=−∫01(∫0∞1+41u2sech2ttsechttanhtdt)du=−π∫01ln(1+41u2+21u)udu=−201π3 using the result of Problem 30.
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Nice. By the way, Taylor expansion of x1arctan(1+x2x) is n=0∑+∞2n+1(−1)nL2n+1x2n
and,
Taylor expansion of x(1+x2)arctan(1+x2x) is n=0∑+∞(−1)n(p=0∑n2p+1L2p+1)x2n
Lk is the k-th Lucas number.
(+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.
Problem 34:
Prove that ∫−∞∞1+x2+x4+⋯+x2k1dx=k+12π⋅sin(2k+23π)cos(2k+2π).
This problem has been solved by Mark Hennings.
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If we define ζ=ek+1πi, then fk(z)=j=0∑kzj=z−1zk+1−1=j=1∏k(z−ζ2j) and hence, using partial fractions, we must be able to write fk(z)1=j=1∑kz−ζ2jAj where Aj=z→ζ2jlimfk(z)z−ζ2j=[fk′(ζ2j)]−1=k+1ζ2j(ζ2j−1) Thus fk(z2)1=j=1∑kz2−ζ2jAj=j=1∑k2ζjAj(z−ζj1−z+ζj1) and hence the function fk(z2)1 has simple poles at ±ζj for 1≤j≤k. Thus ∫−∞∞fk(x2)1dx=2πij=1∑kResz=ζjfk(z2)1=2πij=1∑k2ζjAj=k+1πij=1∑kζj(ζ2j−1)=k+1πi[ζ3−1ζ3(ζ3k−1)−ζ−1ζ(ζk−1)]=k+1πi[ζ3−1ζ3(−ζ−3−1)−ζ−1ζ(ζ−1−1)]=k+1πi[ζ−1ζ+1−ζ3−1ζ3+1]=k+12πi×(ζ−1)(ζ3−1)ζ(ζ2−1)=k+12πi×ζ3−1ζ(ζ+1)=k+12πi×e2(k+1)3πi2isin(2(k+1)3π)e2(k+1)3πi2cos(2(k+1)π)=k+12πsin(2(k+1)3π)cos(2(k+1)π) as required.
Problem 35:
Show that
∫01x2+x+1lnxdx=92[32π2−ψ′(31)]
This problem has been solved by Fdp Dpf.
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∫011+x+x2lnxdx=∫011−x31−xlnxdx=∫011−x31lnxdx−∫011−x3xlnxdx=n=0∑+∞(∫01x3nlnx)−n=0∑+∞(∫01x3n+1lnx)=−n=0∑+∞(3n+1)21+n=0∑+∞(3n+2)21=−n=0∑+∞(3n+1)21+ζ(2)−n=0∑+∞(3n+1)21−n=1∑+∞(3n)21=−2n=0∑+∞(3n+1)21+98ζ(2)=274π2−92n=0∑+∞(n+31)21=274π2−92ψ′(31) Since ψ′(a)=n=0∑+∞(n+a)21
PROBLEM 39:
Evaluate ∫0∞(e−xa−1+xb1)xdx for any a,b>0.
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Solution to problem 39.
J=∫0+∞(e−xa−1+xb1)x1dx
Perform the change of variable y=xa,
J=a1∫0+∞(e−x−1+xba1)x1dx
Perform integration by parts,
aJ=[(e−x−1+xba1)lnx]0+∞+∫0+∞(e−x−ba(1+xba)2xba−1)lnxdx=∫0+∞e−xlnxdx−ba∫0+∞(1+xba)2xba−1lnxdx
∫0+∞(1+xba)2xba−1lnxdx=∫01(1+xba)2xba−1lnxdx+∫1+∞(1+xba)2xba−1lnxdx
In the latter integral perform the change of variable y=x1,
∫0+∞(1+xba)2xba−1lnxdx=∫01(1+xba)2xba−1lnxdx−∫01(1+xba)2xba−1lnxdx=0
∫0+∞e−xlnxdx=Γ′(1)=−γ
Therefore,
J=−a1γ
Problem 43 : Prove That ∫0∞1+(x+tanx)2dx=2π
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Note that,
tanx=−n→∞lim(x−2π1+x+2π1+…+x−(2n−1)2π1+x+(2n−1)2π1)
Also, by Glasser's Master Theorem (which can be proved using elementary methods by using the graphical properties of the function), we have,
∫−∞+∞f(x−x−λ1a1−⋯−x−λnan)dx=∫−∞+∞f(x)dx;ai>0 , λi∈R
Therefore,
∫−∞+∞f(x+atanx)dx=∫−∞+∞f(x)dx;a>0
Putting f(x)=x2+b21, we have,
∫−∞∞(x+atanx)2+b2dx=∫−∞∞x2+b2dx
∴∫0∞(x+atanx)2+b2dx=2bπ□
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I think there are some big convergence issues to be settled The convergence of the series for tanx 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 as the limit of ∫−∞∞f(x−x−λ1a1−⋯−x−λnan)dx
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=∫0∞1+(x+tanx)21dx=21∫R1+(x+tanx)2dx=21n∈Z∑∫−21π21π1+(nπ+x+tanx)21dx=2π21∫−21π21πn∈Z∑(n+πx+tanx)2+π21dx=2π21∫−21π21πcosh2−cos(2x+2tanx)π2sinh2dx=21sinh2∫−21π21πcosh2−cos(2x+2tanx)dx 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=21sinh2∫−21π21πcosh2−cos2xdx=41sinh2∫−ππcosh2−cosxdx Making the complex substitution z=eix, this becomes I=41sinh2∫∣z∣=1cosh2−21(z+z−1)1izdz=−2isinh2∫∣z∣=1(z−e2)(z−e−2)dz=−πsinh2Resz=e−2(z−e2)(z−e−2)1=21π as required.
Problem 38:
Evaluate,
∫04πcosx(cosx+sinx)xln(cosx−sinxcosx+sinx)dx
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We note that I=∫041πcosx(cosx+sinx)xln(cosx−sinxcosx+sinx)dx=∫041π1+tanxxln(1−tanx1+tanx)sec2xdx=∫011+utan−1uln(1−u1+u)du=−∫01tan−1(1+y1−y)1+ylnydy after the substitutions u=tanx and y=1+u1−u. Since tan(41π−tan−1y)=1+y1−y we deduce that I=−∫01(41π−tan−1y)1+ylnydy=−41π∫011+ylnydy+∫011+ytan−1ylnydy Now it is elementary that ∫011+ylnydy=−121π2 and we see from this MSE post --- hello, it's one of yours again --- that ∫011+ytan−1ylnydy=21Gln2−641π3 so it follows that I=1921π3+21Gln2
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Congrats ! It's like you read my mind, you know all my tricks ;) Happy new year to all !
Problem 40
Evaluate ∫0∞2x−1 ln(2x−1)x−1dx
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Substitute (2x−1)=t2 to get,
I=ln221∫0∞((t2+1)lntln(t2+1)−ln2)dt
Substitute t↦t1
⟹I=−ln221∫0∞((t2+1)lntln(t2+1)−ln2)dt+ln222∫0∞t2+1dt
⟹I=−I+ln22π
⟹I=2ln22π
PROBLEM 42 :
Evaluate ∫01(x−1)lnx(xp−1)(xq−1)dxp,q>0
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Let,
I=∫01(x−1)lnx(xp−1)(xq−1) dx
=−∫01(1−x)lnx(xp−1)(xq−1) dx
=−r=0∑∞∫01[(lnxxp+q+r−xp+r)−(lnxxq+r−xr)] dx
=−n→∞limr=0∑n[log(p+r+1p+q+r+1)−log(r+1q+r+1)]
=−n→∞limr=1∑n[log(p+rp+q+r)−log(rq+r)]
where the standard result ∫01lnxxa−xb dx=ln(b+1a+1) has been used in the above lines.
⟹I=−n→∞limlog([(p+1)(p+2)…(p+n)]×[(q+1)(q+2)…(q+n)]n!×(p+q+1)(p+q+2)…(p+q+n))
=−n→∞limlog(Γ(p+q+1)Γ(p+n+1)Γ(q+n+1)Γ(p+q+n+1)Γ(p+1)Γ(n+1)Γ(q+1))
Since n→∞limΓ(n+x+1)nxΓ(n+1)=1, we have,
I=log(Γ(p+1)Γ(q+1)Γ(p+q+1))□
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One of the standard integral representations of lnΓ(z) gets you there much more easily...
Problem 33:
Evaluate ∫041πcosp+2xsinpxdxp>0.
This problem has been solved by Sumanth R Hegde.
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Put tanx=t
The integral reduces to 0∫1tpdt
This gives p+11
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Problem 41 :
Prove That
∫−∞∞(xn) dx=2nℜ(n)>−1
Notation : (xn) denotes the Generalized Binomial Coefficient.
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Presumably, we mean (xn)=Γ(x+1)Γ(n+1−x)n!x∈R Now Γ(x+1)Γ(n+1−x)=xΓ(x)j=1∏n(j−x)Γ(1−x)=(−1)nsinπxπ×j=0∏n(x−j) and so we can use partial fractions to write (xn)=π(−1)nn!sinπx(j=0∏n(x−j))−1=π(−1)nn!sinπxj=0∑nx−jAj where Aj=j!(n−j)!(−1)n−j0≤j≤n Since ∫Rx−jsinπxdx=(−1)j∫Rxsinπxdx=(−1)jπ we deduce that ∫R(xn)dx=π(−1)nn!j=0∑nAj∫Rx−jsinπxdx=(−1)nn!j=0∑n(−1)jAj=j=0∑nj!(n−j)!n!=j=0∑n(jn)=2n as required.
If we insist on being picky, we could integrate from −X to X, and let X→∞, to avoid dealing with the improper integral ∫Rxsinxdx directly. We only have to combine a finite number of integrals at the last stage, so there is no problem taking the limit.
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The integral is also true for ℜ(n)>−1.
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∫−∞∞Γ(α+x)Γ(β−x)dx=Γ(α+β−1)2α+β−2 for Re(α+β)>1 is 6.414.2 in Gradshteyn & Rhyzik.
Indeed. The formulaeIsn't it a bit late to be altering the question, though?
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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))m−1einx dx=2m−1π mB(21(m+n+1),21(m−n+1))ein/2
If (x+1n+1)=(xn)+(x+1n) then i think it's straightforward to prove with a recurrence proof.
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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.
Problem 45.
Evaluate ∫0∞sinh(πx)(1+x6)x5dx
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Using the Fourier transform we have I=∫0∞sinhπx(1+x6)x5dx=21∫−∞∞sinhπx(1+x6)x5dx=21⟨f,g⟩=21⟨Ff,Fg⟩ where f(x)=sinhπxxg(x)=1+x6x4 A standard result about Fourier transforms tells us that (Ff)(u)=22π1sech221u and the Fourier transform of g can be readily calculated using contour integration: for u>0 we simply need to determine the residues of 1+z6z4eizu at the three roots of z6+1=0 with positive imaginary part. Omitting the details, (Fg)(u)=312π[e−∣u∣+e−21∣u∣cos(23∣u∣)−3e−21∣u∣sin(23∣u∣)] Thus we deduce that I=21⟨Ff,Fg⟩=121∫0∞sech221u[e−u+e−21ucos(23u)−3e−21usin(23u)]du Now ∫1∞(1+v)2vadu=21(1−aH−21−2a+aH−21a)Rea<1 and so ∫0∞eωusech221udu=4∫0∞(e21u+e−21u)2eωudu=4∫1∞(1+v)2vωdv=2(1−ωH21ω2+ωH−21ω) where ω is the primitive cube root of unity. Taking real and imaginary parts appropriately, and doing a lot of simplification with polygammas, ∫0∞e−21u[cos(23u)−3sin(23u)]sech221udu=−2+4πsech(23π) On the other hand, ∫0∞e−usech221udu=4∫1∞v(v+1)2dv=−2+4ln2 Putting this all together, we obtain that I=31[−1+ln2+πsech(23π)] Someone else can post the next one, please.
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Here is a partial progress towards an alternate solution f(a)=∫0∞sinh(πx)(1+x6)sinh(ax) dx
then,
f(6)(a)+f(a)=21tan(2a)
where f(n)(a) denotes the nth derivative of f w.r.t. a, with f(0)=0 and the required integral being f(5)(0).
I still have to solve that D.E.
* PROBLEM 46: *
Show that ∫01x4+x2+11−x2dx=2243π(3−1)
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Perform the change of variable y=1+x1−x,
∫01x4+x2+11−x2dx=∫013⋅x4+10⋅x2+3x⋅(4+4⋅x)dx=[(223⋅343−223⋅345)⋅atan(2⋅3412⋅y−2⋅341)+(223⋅343−223⋅345)⋅atan(2⋅3412⋅341+2⋅y)+(223⋅343−223⋅345)⋅atan(2⋅3412⋅3⋅y−2⋅341)+(223⋅343−223⋅345)⋅atan(2⋅3412⋅341+2⋅3⋅y)+(2⋅345+2⋅343)⋅log(y−2⋅341⋅y+3)+(−2⋅345−2⋅343)⋅log(y+2⋅341⋅y+3)+(−2⋅345−2⋅343)⋅log(3⋅y−2⋅341⋅y+1)+(2⋅345+2⋅343)⋅log(3⋅y+2⋅341⋅y+1)×−241]01=4⋅341(2−2⋅3)⋅atan(341341−2)+(2⋅3−2)⋅atan(3412+341)+(2⋅3−2)⋅atan(2⋅341−1)+(2⋅3−2)⋅atan(2⋅341+1)=4×341(6−2)(arctan(3+23)+π−arctan(3+23))=2243π(3−1)
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My approach used contour integration...
The function z−1 can be defined analytically on the cut plane C\{(−∞,1], while the function z+1 can be defined analytically on the cut plane C\(−∞,−1]. Putting these two functions together, it is possible to define the function z2−1 analytically on the domain C\[−1,1].
Consider the "dogbone" contour Dε=−γ1+γ2+γ3+γ4, where
Then we deduce that ε→0+lim∫Dεz4+z2+1z2−1dz=−2i∫−11x4+x2+11−x2dx=−4i∫01x4+x2+11−x2dx If we let ΓR be the circular contour z=Reiθ for 0≤θ<2π, then we have (∫ΓR−∫Dε)z4+z2+1z2−1dz=2πiu∈{±ω,±ω2}∑Resz=uz4+z2+1z2−1 for sufficiently small ε>0 and any R>1. Letting R→∞ and ε→0+, we deduce that ∫01x4+x2+11−x2dx=21πu∈{±ω,±ω2}∑Resz=uz4+z2+1z2−1=21πu∈{±ω,±ω2}∑Resz=uz6−1(z2−1)z2−1=121πu∈{±ω,±ω2}∑u(u2−1)u2−1 If u=eiθ for ∣θ∣<π, then u+1u−1u(u2−1)u+1u−1u2−1u(u2−1)u2−1=2cos21θe21iθ=2isin21θe21iθ=2isinθe2iθ=2cos21θe41iθ=2∣sin21θ∣e41i(θ+πsgn(θ))=2∣sinθ∣e41i(2θ+πsgn(θ))=2isinθ2∣sinθ∣e41i(10θ+πsgn(θ)) and hence ∫01x4+x2+11−x2dx=121π×4×343sin121π=341πsin121π=223341π(3−1)
Problem 47:
Evaluate,
∫011−x2arctanxdx
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The sequence of substitutions u=x2, v=1−u, w=v, y=sinhp and q=ep give I=∫011−x2tan−1xdx=∫011−x2dx∫011+x2y2xdy=∫01dy∫011−x2(1+x2y2)xdx=21∫01dy∫011−u(1+y2u)du=21∫01dy∫01v(1+y2−y2v)dv=∫01dy∫011+y2−y2w2dw=21∫01ln(1+y2−y1+y2+y)y1+y2dy=∫0sinh−11sinhppdp=2∫0sinh−11e2p−1pepdp=2∫11+2q2−1lnqdq=−[lnqln(q+1)+Li2(−q)+Li2(1−q)]11+2=−ln(1+2)ln(2+2)−Li2(−1−2)−Li2(−2)−121π2 We now use some dilogarithm identities Li2(2)+Li2(21)Li2(−2)Li2(−1−2)+Li2(21)=31π2−21ln22−iπln2=31π2−81ln22−21πiln2=21Li2(2)−Li2(2)=21(41π2−πiln2)−31π2+81ln22+21πiln2+Li2(21)=−245π2+81ln22+Li2(21)=−21ln2(2+2) to deduce that Li2(−1−2)+Li2(−2)=−245π2+81ln22−21ln2(2+2) and hence that I=−ln(1+2)ln(2+2)+245π2−81ln22+21ln2(2+2)−121π2=81π2−21ln2(1+2)
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J=∫011−x2arctanxdx
Perform the change of variable y=1+x1−x,
J=2∫011+x2arctan(1+x21−x2)dx=2∫011+x2arctan(1)dx−2∫011+x2arctan(x2)dx=8π2−2∫011+x2arctan(x2)dx
∫011+x2arctan(x2)dx=[arctanxarctan(x2)]01−∫011+x42xarctanxdx=16π2−∫011+x42xarctanxdx
Since,
arctanx=∫011+t2x2xdx
then,
K=∫011+x42xarctanxdx=∫01∫01(1+t2x2)(1+x4)2x2dtdx=∫01∫01((1+t4)(1+x4)2t2+(1+x4)(1+t4)2x2−(1+t4)(1+t2x22t2)dtdx=4(∫011+t4t2dt)(∫011+x41dx)−K
Therefore,
K=2(∫011+x4x2dx)(∫011+x41dx)
Since,
∫011+x4x2dx=[421ln(x2+2x+1x2−2x+1)+221arctan(2x+1)+221arctan(2x−1)]01=421(π+ln(3−22))
and,
∫011+x41dx=[421ln(x2−2x+1x2+2x+1)+221arctan(2x+1)+221arctan(2x−1)]01=421(π−ln(3−22))
Therefore,
K=16π2−161(ln(3−22))2
Therefore,
∫011+x2arctan(x2)dx=161(ln(3−22))2
Therefore,
J=8π2−81(ln(3−22))2
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y=1+x1−x come from?
You do like pulling substitutions out of nowhere! Where didThis is a nice alternative derivation. Of course, since 3−22=(1+2)−2, our results are the same!
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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.
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∫01 i always try the change of variable y=1+x1−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=x. My tool to do such computations is Maxima.
To respond your question. When i have an integral* PROBLEM 44 *:
Evaluate ∫01ln(1−ax1+ax)x1−x2dx−1<a<1.
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For −1<ax<1, log(1−ax1+ax)=2n=0∑∞(2n+1)a2n+1x2n+1
∫01x1−x2log(1+ax1−ax)dx=2∫01(n=0∑∞(2n+1)a2n+1x2n+1⋅x1−x21)dx=2n=0∑∞(∫01(2n+1)a2n+1x2n+1⋅x1−x21dx)=2n=0∑∞(2n+1a2n+1∫011−x2x2ndx)=n=0∑∞2n+11⋅B(21,n+21)a2n+1=n=0∑∞2n+11Γ(n+1)Γ(21)Γ(n+21)a2n+1=n=0∑∞2n+11n!22nn!(2n)!πa2n+1=πn=0∑∞22n(2n+1)(n2n)a2n+1=πarcsin(a)
For variety, if we define F(a)=∫01ln(1−ax1+ax)x1−x2dx−1<a<1 then F′(a)=∫01(1+ax1+1−ax1)1−x2dx=∫021π(1+acosθ1+1−acosθ1)dθ=∫0π1+acosθdθ=21∫02π1+acosθdθ=21∫∣z∣=12+a(z+z−1)2izdz=−i∫∣z∣=1az2+2z+adz=−ia∫∣z∣=1(az+1+1−a2)(az+1−1−a2)dz=2πaResz=a−1(1−a2−1)(az+1+1−a2)(az+1−1−a2)1=1−a2π and hence F(a)=∫0a1−b2πdb=πsin−1a
* PROBLEM 48: *
Prove that ∫0∞(x+r)μ(x+s)μxμ−21dx=π(r+s)1−2μΓ(μ)Γ(μ−21) for all r,s>0 and μ>21.
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Someone else can post the next one...
For any a>0 and 0<b<c and ∣x∣<1 we have Ia,b,c(x)=∫01tb−1(1−t)c−b−1(1−xt)−adt=∫01tb−1(1−t)c−b−1(n=0∑∞n!(a)(n)xntn)dt=n=0∑∞n!(a)(n)xn∫01tn+b−1(1−t)c−b−1dt=n=0∑∞n!(a)(n)xnB(n+b,c−b)=n=0∑∞n!(a)(n)xnΓ(n+c)Γ(n+b)Γ(c−b)=n=0∑∞n!(a)(n)xn(c)(n)Γ(c)(b)(n)Γ(b)Γ(c−b)=B(b,c−b)n=0∑∞(c)(n)(a)(n)(b)(n)n!xn=B(b,c−b)(2F1)(a,b;c;x) With the substitution y=1−tt we see that ∫0∞yb−1(1+y)a−c(1+αy)−ady=∫01(1−tt)b−1(1−t1)a−c(1−t1−(1−α)t)−a(1−t)2dt=∫01tb−1(1−t)c−b−1(1−(1−α)t)−adt=Ia,b,c(1−α)=B(b,c−b)(2F1)(a,b;c;1−α) for any a>0, 0<b<c and 0<α≤1.
Since the integral is symmetric in r and s, we may assume that 0<r≤s. Then, for any 0<r≤s and μ>21 we have Jμ,r,s=∫0∞(x+r)μ(x+s)μxμ−21dx=sμr∫0∞tμ−21(1+t)−μ(1+srt)−μdt=sμrB(μ+21,μ−21)(2F1)(μ,μ+21;2μ;1−sr) Since (2F1)(a,b;c;z)=(1−z)−aΓ(b)Γ(c−a)Γ(c)Γ(b−a)(2F1)(a,c−b;a−b+1;(1−z)−1)+(1−z)−bΓ(a)Γ(c−b)Γ(c)Γ(a−b)(2F1)(b,c−a;b−a+1;(1−z)−1) we deduce that Jμ,r,s=sμrB(μ+21,μ−21)⎣⎢⎢⎢⎡(sr)−μΓ(μ+21)Γ(μ)Γ(2μ)Γ(21)(2F1)(μ,μ−21;21;rs)+(sr)−μ−21Γ(μ)Γ(μ−21)Γ(2μ)Γ(−21)(2F1)(μ+21,μ;23;rs)⎦⎥⎥⎥⎤=rμ22μ−1B(μ+21,μ−21)[r(2F1)(μ,μ−21;21;rs)−s(2μ−1)(2F1)(μ+21,μ;23;rs)]=rμπΓ(μ)Γ(μ−21)[r(2F1)(μ,μ−21;21;rs)−s(2μ−1)(2F1)(μ+21,μ;23;rs)] Now standard hypergeometric identities give (2F1)(μ,μ−21;21;rs)(2F1)(μ,μ+21;23;rs)=21[(1−rs)1−2μ+(1+rs)1−2μ]=2(2μ−1)1sr[(1−rs)1−2μ−(1+rs)1−2μ] and so Jμ,r,s=rμ−21πΓ(μ)Γ(μ−21)(1+rs)1−2μ=πΓ(μ)Γ(μ−21)(r+s)1−2μ as required.
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Nice. I'm working on an alternative solution. I hope to get something soon.
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!
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It makes things easier :)
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Please don't post the answer now, post the solution at the end of the week end. Thanks.
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I was aiming to allow some extra time to compensate for the typo, but will post a solution tomorrow...
Problem 49:
Evaluate,
∫06π4sin4x−2sin2x+1xsin(2x)dx
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For any a>0 define A(a)B(a)a4A(a)B(a)=∫01a4+x4dx=a31∫0a−11+z4dz=2a321[tan−1(1+2a−1)−tan−1(1−2a−1)]+4a321ln(a−2+a−1a+2+a−1)=∫01a4+x4x2dx=a1∫0a−11+z4z2dz=2a21[tan−1(1+2a−1)−tan−1(1−2a−1)]−4a21ln(a−2+a−1a+2+a−1)=81[tan−1(1+2a−1)−tan−1(1−2a−1)]2−321ln2(a−2+a−1a+2+a−1) Thus we deduce that (putting z=ax and playing symmetry games) I(a)=∫0a−11+x4xtan−1(ax)dx=∫0a−11+x4xdx∫011+a−2x2y2axdy=a4∫01∫01(a4+z4)(a4+y2z2)z2dydz=21a4∫01∫01{(a4+z4)(a4+y2z2)z2+(a4+y4)(a4+y2z2)y2}dydz=21a4∫01∫01(a4+y4)(a4+z4)y2+z2dydz=a4A(a)B(a)=81[tan−1(1+2a−1)−tan−1(1−2a−1)]2−321ln2(a−2+a−1a+2+a−1) Moreover J(a)=∫0a−1x2+a2tan−1x2dx=[a1tan−1x2tan−1(ax)]0a−1−a2∫0a−11+x4xtan−1(ax)dx=a1(tan−1a−2)2−a2I(a) Thus, putting u=cos2x and v=32u−1, we see that K=∫061π4sin4x−2sin2x+1xsin2xdx=∫061πcos22x−cos2x+1xsin2xdx=41∫211u2−u+1cos−1udu=41[32tan−1(32u−1)cos−1u]211+41∫21132tan−1(32u−1)1−u2du=231∫211tan−1(32u−1)1−u2du=2×3411∫031(1−3v)(3+v)tan−1vdv The key substitution w=3+v1−3v yields K=3411∫03−41tan−1(3+w21−3w2)3+w2dw=3411∫03−41[61π−tan−1w2]3+w2dw=6×341π[3−41tan−1(341w)]03−41−3−41J(341)=363π2−3−41[3−41(tan−13−21)2−2×3−41I(341)]=32I(341)=431[tan−1(341−3−412)]2−1631ln2(341−2+3−41341+2+3−41) noting that tan−1(1+2a−1)−tan−1(1−2a−1)=tan−1(a−a−12)a>1
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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.
Actually my solution is the same than yours. The only difference is the way you manage to transform the trigonometric integral.
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32I(31/4) (in my notation), is elegant.
Yes, but your method of transforming the original integral into what is, apart from a variable scaling,Log in to reply
F(a,t,x)=(t⋅x+a)⋅(x2+a)x, compute ∫01F(1,t,x)dt isn't nice?
ifProblem 47 relies on the evaluation of ∫01∫01F(1,t2,x2)dtdx and problem 49 relies on ∫01∫01F(3,t2,x2)dtdx
Who will post problem 50, the last one?
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32I(341). What I was saying was that your technique for converting the original integral to 32I(341) was more elegant than mine - in Mathematics, the word "elegant" is complimentary!
Let me clarify. Both of our methods evaluateAliter,
Let.
I=∫06π4sin4x−2sin2x+1xsin(2x) dx
=4(α−β)1∫06πxsin(2x)(cos(2x)−1+β1−cos(2x)−1+α1)
where α and β are the roots of the equation t2−21t+41=0
Now, from Problem 25, we have,
p2+2pcosx+1sinx=k=1∑∞(−p)k−1sin(kx)
Using this and interchanging, sum and integral ( and after a lot of straight forward simplification/calculations involving dilogarithm and logarithm) we get,
I=431[tan−1(341−3−412)]2−1631ln2(341−2+3−41341+2+3−41)
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Just pipped you! You might want to clarify what α and β are, though.
Why don't you post the last integral!
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What was your method?
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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.
J=∫06π1−2sin2x+4sin4xxsin(2x)dx=∫06π(1−sin2x)2+3sin4x2xsinxcosxdx=∫06πcos4+3sin4x2xsinxcosxdx=∫06πcos2x(1+3tan4x)2xtanxdx
Perform the change of variable y=3tanx,
J=∫013+x42xarctan(3x)dx
Since,
arctanu=∫011+t2u2udt
J=23∫01∫01(3+x4)(3+t2x2)x2dtdx=23∫01∫01((t4+3)(x4+3)t2+x2−(t4+3)(t2x2+3)t2)dtdx=43∫01∫01(3+t4)(3+x4)t2dtdx−J
therefore,
J=23∫01∫01(3+t4)(3+x4)t2dtdx=23(∫013+x4x2dx)(∫013+x41dx)
Since,
∫013+x41dx=⎣⎢⎢⎡223341arctan(3−x22⋅341x)−225341log(x2+2⋅341x+33−2⋅341x+x2)⎦⎥⎥⎤01=2233431arctan(22⋅341+22⋅343)−2253431log(1+3+2⋅3411+3−2⋅341)
1+3+3⋅341 is a root of the polynomial X4−4X3−16X+16,
Therefore,
∫013+x41dx=2233431arctan(3−12⋅341)−2253431log(1+3−22341−22343)
and,
∫013+x4x2dx=⎣⎢⎢⎡223⋅341arctan(3−x22⋅341⋅x)+225⋅341log(x2+2⋅341⋅x+33−2⋅341⋅x+x2)⎦⎥⎥⎤01=2233411arctan(22⋅341+22⋅343)+2253411log(1+3−22341−22343)
Therefore,
J=431arctan2(22⋅341+22⋅343)−1631log2(1+3−22341−22343)
PS:
F(x,t)=(a+x2)(a+tx)x=a+t2t⋅a+x21+a+t21⋅a+x2x−(a+t2)(a+tx)t
is a nice function.
Ohh!! I missed this season. I even didn't know that this happened. :(
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Maybe this time you can host it :)
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When is starting the season 4? :)
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Problem 50 :
Evaluate ∫0∞(1+a2x2)sinxsin(2nx)cos(2arctan(ax)) dx
where a>0 , n∈Z+
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Suppose that a>0 and n∈N. Note that I=∫0∞(1+a2x2)sinxsin(2nx)cos(2tan−1(ax))dx=∫0∞(1+a2x2)21−a2x2sinxsin2nxdx=δ→0+limIδ where Iδ=∫0∞(1+a2(x+δ)2)21−a2(x+δ)2sinxsin2nxdx where we can use the Dominated Convergence Theorem to justify the limit, since ∣∣∣∣(1+a2(x+δ)2)21−a2(x+δ)2sinxsin2nx∣∣∣∣≤∣∣∣∣1+a2(x+δ)21sinxsin2nx∣∣∣∣≤1+a2x2Nn for all x≥0 and δ>0, where ∣∣sinxsin2nx∣∣≤Nn for all x≥0. It is trivial that ∫0∞tcos(at)e−xtdt=a2(1+a2x2)2a2x2−1a,x>0 and so Iδ=−a−2∫0∞(∫0∞tcos(at)e−(x+δ)tdt)sinxsin2nxdx=−a−2∫0∞tcos(at)e−δt(∫0∞e−xtsinxsin2nxdx)dt Since ∫0∞e−xtsinxsin(2m+2)xdx−∫0∞e−xtsinxsin2mxdx=2∫0∞e−xtcos(2m+1)xdx=t2+(2m+1)22t a simple induction shows that ∫0∞e−xtsinxsin2nxdx=2tk=0∑n−1t2+(2k+1)21t>0 and hence Iδ=−2a−2∫0∞cos(at)e−δt(k=0∑n−1t2+(2k+1)2t2)dt=−2a−2∫0∞cos(at)e−δt(n−k=0∑n−1t2+(2k+1)2(2k+1)2)dt=−2a−2[δ2+a−2nδ−k=0∑n−1(2k+1)2∫0∞t2+(2k+1)2cos(at)e−δtdt] and hence, letting δ→0+, I=2a−2k=0∑n−1(2k+1)2∫0∞t2+(2k+1)2cos(at)dt=2a−2k=0∑n−1(2k+1)2×2(2k+1)πe−a2k+1=a2πk=0∑n−1(2k+1)e−a2k+1 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.
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Nice! I used the expansion 2sinxsin(2nx)=k=1∑ncos((2k−1)x)
The result can also be generalized to ∫0∞(1+a2x2)2psinxsin(2nx)cos(parctan(ax)) dx ; n∈Z,a,p>0
Let Cn−1(a) be the previous integral. I suspect one can find a closed form for the ordinary generating function for the sequence (Cn(a)) because U2n+1(cosx)=sinxsin((2n+2)x, Uj being the j-nth Chebyshev polynomial of the second kind.
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I looked hard on the web and could not find a closed expression for the Laplace transform of the Un, which would have been really useful!