Solution for previous proposal

Hello Brilliants, here I wish to share my solution for the proposed problem.

Before we start with the problem we shall consider the integral of the form for all n≥1n\geq 1 I(n)=∫01xn−1ln⁡(1−x)dx=1n∫01ddx(xn−1)ln⁡(1−x)dxI(n) = \displaystyle \int_0^1 x^{n-1} \ln(1-x) dx =\frac{1}{n}\displaystyle \int_0^1 \frac{d}{dx}(x^{n}-1)\ln(1-x) dx and by integration by parts we yield I(n)=1n[(xn−1)ln⁡(1−x)⏟0]01−1n∫011−xn1−xdxI(n)=\frac{1}{n}\left[\underbrace{(x^{n}-1)\ln(1-x)}_{0}\right]_0^1-\frac{1}{n}\int_0^1\frac{1-x^{n}}{1-x}dx =−1n∫01∑j=1nxj−1=−1n∑j=1n∫01xj−1dx=−1n∑j=1n1j=−Hnn⋯(1)=-\frac{1}{n}\int_0^1\sum_{j=1}^{n} x^{j-1} =-\frac{1}{n}\sum_{j=1}^{n}\int_0^1 x^{j-1}dx=-\frac{1}{n} \sum_{j=1}^n\frac{1}{j} =-\frac{H_n}{n}\cdots (1) Note that D(k)=1n2m+m2n+kmn=1mn∫01xm+n+k−1dx=xk−1mn∫01xm+ndxD(k)=\displaystyle \frac{1}{n^2m+m^2n+kmn} = \frac{1}{mn}\int_0^1 x^{m+n+k-1} dx=\frac{x^{k-1}}{mn}\int_0^1{x^{m+n}}dx and therefore ∑1≤n≤m≤∞D(k)=∫01xk−1∑n=0∞∑m=0∞xnn⋅xmmdx=∫01xk−1ln⁡2(1−x)dx\sum_{1\leq n\leq m\leq \infty} D(k)=\int_0^1 x^{k-1} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \frac{x^n}{n}\cdot \frac{x^m}{m}dx =\int_0^1 x^{k-1}\ln^2(1-x) dx Now we consider the latter integral I(k)=∫01xkln⁡2(1−x)dx I(k)=\displaystyle \int_0^1 x^{k}\ln^2(1-x) dx for k≥0k\geq 0 which further can be written as 1k+1∫01ddx(xk+1−1)ln⁡2(1−x)dx\displaystyle \frac{1}{k+1}\int_0^1 \frac{d}{dx}(x^{k+1}-1)\ln^2(1-x) dx and hence on Integration by parts we see that I(k)=1k+1[(xk+1−1)ln⁡2(1−x)⏟0]01−2k+1∫011−xk+11−xln⁡(1−x)dxI(k)=\frac{1}{k+1}\left[\underbrace{(x^{k+1}-1)\ln^2(1-x)}_{0}\right]_0^1-\frac{2}{k+1}\int_0^1\frac{1-x^{k+1}}{1-x}\ln(1-x)dx =−2k+1∫01∑n=1k+1xn−1ln⁡(1−x)=−2k+1∑n=1k+1∫01xn−1ln⁡(1−x)dx=-\frac{2}{k+1}\int_0^1\sum_{n=1}^{k+1} x^{n-1}\ln(1-x)=-\frac{2}{k+1}\sum_{n=1}^{k+1}\int_0^1 x^{n-1}\ln(1-x)dx Plugging the result from (1)(1) to last integral we have 2k+1∑n=1k+1Hnn=2k+1(Hk+12+Hk+1(2)2)=Hk+12+Hk+1(2)k+1\frac{2}{k+1}\sum_{n=1}^{k+1}\frac{H_n}{n}=\frac{2}{k+1}\left(\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2}\right)=\frac{H_{k+1}^2 +H_{k+1}^{(2)}}{k+1} Further note that nth partial sum of Hk+1(2)=ζ(2)−ψ1(k+2)H_{k+1}^{(2)} = \zeta(2) -\psi^1(k+2) giving us required result of D(k)=Hk+12−ψ1(k+2)k+1+π26(k+1)D(k)=\frac{H_{k+1}^2-\psi^1(k+2)}{k+1}+\frac{\pi^2}{6(k+1)}

Now we notice that 1(q+b)q+b+1+(q+b+1)q+b=(q+b)q+b+1−(q+b+1)q+b(q+b)q+b+1(−1)\frac{1}{(q+b)\sqrt{q+b+1}+(q+b+1)\sqrt{q+b}}=\frac{(q+b)\sqrt{q+b+1}-(q+b+1)\sqrt{q+b}}{(q+b)\sqrt{q+b+1}(-1)}=1q+b−1q+b+1=\frac{1}{\sqrt{q+b}}-\frac{1}{\sqrt{q+b+1}} and hence ∑n=0∞∑q=0nxn(1q+b−1q+b+1⏟Telescoping)=∑n=0∞xn(1b−1n+b+1)\sum_{n=0}^{\infty}\sum_{q=0}^n x^n\left(\underbrace {\frac{1}{\sqrt{q+b}}-\frac{1}{\sqrt{q+b+1}}}_{\text{Telescoping}}\right)=\sum_{n=0}^{\infty}x^n\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{n+b+1}}\right) and by hence by algebra of sum and for for ∣x∣<1|x|<1 ∑n=0∞xnb−∑n=0∞xnn+b+1=1b(1−x)−Φ(x,12,1)\sum_{n=0}^{\infty}\frac{x^n}{\sqrt{b}}-\sum_{n=0}^{\infty}\frac{x^n}{\sqrt{n+b+1}}=\frac{1}{\sqrt{b}(1-x)}-\Phi\left(x, \frac{1}{2},1\right) and we deduce that α=2,β=1,γ=6 \alpha = 2, \beta =1 ,\gamma =6. Thus Φ(β−α,β,(α+2βγ)−1)=Φ(−1,1,110)=∑n=0∞(−1)n(n+110)=D\Phi\left(\beta-\alpha,\beta , (\alpha+2\beta \gamma)^{-1}\right)=\Phi\left(-1,1,\frac{1}{10}\right)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{\left(n+\frac{1}{10}\right)}=D D=∑n=0∞1(2n+10−1)(2n+10−1+1)=14∑n=0∞1(n+20−1)(n+20−1+2−1)D= \sum_{n=0}^{\infty}\frac{1}{\left(2n+10^{-1}\right)(2n+10^{-1}+1)}=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{(n+20^{-1})(n+20^{-1}+2^{-1})} =2−1(ψ(12+110)−ψ(120))=2−1(ψ(110)−2ψ(120)−2ln⁡2)=2^{-1}\left(\psi\left(\frac{1}{2}+\frac{1}{10}\right)-\psi\left(\frac{1}{20}\right)\right)=2^{-1}\left(\psi\left(\frac{1}{10}\right)-2\psi\left(\frac{1}{20}\right)-2\ln 2\right)and finally we have D=ψ(110)−ψ(120)−ln⁡2\displaystyle D=\psi\left(\frac{1}{10}\right)-\psi\left(\frac{1}{20}\right)-\ln 2. Here to evaluate this expression DD , we use the Gauss's-Diagamma theorem ie ψ(rm)=−γ−ln⁡(2m)−π2cot⁡(rπm)+2∑q=1⌊m−12⌋cos⁡(2πqrm)ln⁡sin⁡(πqn)\psi\left(\frac{r}{m}\right)=-\gamma -\ln(2m)-\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right)+2\sum_{q=1}^{\left\lfloor \frac{m-1}{2}\right\rfloor}\cos\left(\frac{2\pi qr}{m}\right)\ln\sin\left(\frac{\pi q}{n}\right) plugging r=1,m=10,20 r=1, m=10 ,20 we have I=ψ(110)=−γ−ln⁡(20)−π2cot⁡(π10)+2∑q=14cos⁡(πq5)ln⁡sin⁡(πq10)I= \psi\left(\frac{1}{10}\right)=-\gamma -\ln(20)-\frac{\pi}{2}\cot\left(\frac{\pi}{10}\right)+2\sum_{q=1}^{4}\cos\left(\frac{\pi q}{5}\right)\ln\sin\left(\frac{\pi q}{10}\right) and K=ψ(120)=−γ−ln⁡(40)−π2cot⁡(π20)+2∑q=19cos⁡(πq10)ln⁡sin⁡(πq20)K= \psi\left(\frac{1}{20}\right)=-\gamma -\ln(40)-\frac{\pi}{2}\cot\left(\frac{\pi}{20}\right)+2\sum_{q=1}^{9}\cos\left(\frac{\pi q}{10}\right)\ln\sin\left(\frac{\pi q}{20}\right) and also D=I−k−ln⁡2D=I-k -\ln 2 and on further simplification of the DD we obtain that D=π41sin⁡(120)cos⁡(120)+A(log⁡cot⁡(π20))+B(log⁡cot⁡(3π20))D = \frac{\pi}{4}\frac{1}{\sin\left(\frac{1}{20}\right)\cos\left(\frac{1}{20}\right)}+A\left(\log\cot\left(\frac{\pi}{20}\right)\right)+B\left(\log\cot\left(\frac{3\pi}{20}\right)\right) with A=5+52=ϕ+2A=\sqrt{\frac{5+\sqrt 5}{2}}=\sqrt{\phi+2} and B=5−52=2−110−25B=\sqrt{\frac{5-\sqrt{5}}{2}}=2^{-1}\sqrt{10-2\sqrt{5}} and π41sin⁡(π20)cos⁡(π20)=π21sin⁡(18∘)=2π5−1\displaystyle \frac{\pi}{4}\frac{1}{\sin\left(\frac{\pi}{20}\right)\cos\left(\frac{\pi}{20}\right)}\\ =\frac{\pi}{2}\frac{1}{\sin(18^{\circ})}=\frac{2\pi}{\sqrt 5-1} since sin⁡(18∘)=4−1(5−1)\sin(18^{\circ} )=4^{-1}\left(\sqrt 5-1\right).

Further we have cot⁡(π20)=cot⁡(9∘)\displaystyle\cot\left(\frac{\pi}{20}\right)=\cot(9^{\circ}). To evaluate cot⁡(9∘\cot(9^{\circ} we shall be using the fact that cot⁡(3x)=cos⁡(3x)sin⁡(3x)=4cos⁡3x−3cos⁡x3sin⁡x−4sin⁡3x=cos⁡x(1−4sin⁡2x)sin⁡x(3−4sin⁡2x)=cot⁡x(1−23−4sin⁡2x)\cot (3x)=\frac{\cos (3x)}{\sin (3x)}= \frac{4 \cos^3 x-3\cos x}{3\sin x-4\sin^3 x}=\frac{\cos x(1-4\sin ^2 x)}{\sin x(3-4\sin^2 x)}=\cot x\left(1-\frac{2}{3-4\sin^2 x}\right) setting x=3x=3 we have the expression for cot⁡9∘\cot9^{\circ} but then we will show that sin⁡3∘=148−10−25−15−3\sin 3^{\circ}=\frac{1}{4}\sqrt{8-\sqrt{10-2\sqrt 5}-\sqrt{15}-\sqrt{3}} {Note}: This identity is proposed by Narendra Bhandari and is proved by Sergio Esteban beautiful geometry work) and Ahmed Hegazi(trigonometry work). From half angle formula we have that sin⁡ϕ2=1−cos⁡ϕ2\sin\frac{\phi}{2}=\sqrt{\frac{1-\cos \phi}{2}} now set ϕ=6∘\phi = 6^{\circ}. Giving us sin⁡3∘=1−cos⁡6∘2=1−cos⁡(36∘−30∘)2\sin3^{\circ} =\sqrt{\frac{1-\cos 6^{\circ}}{2}}= \sqrt{\frac{1-\cos(36^{\circ} -30^{\circ})}{2}} Now using compound angle formula we can get
cos⁡(36∘−30∘)=cos⁡36∘cos⁡30∘+sin⁡36∘sin⁡30∘=(5+14)32+(10−254)12=10−25+3+158\displaystyle \cos(36^{\circ} -30^{\circ}) = \cos36^{\circ}\cos30^{\circ}+\sin36^{\circ}\sin30^{\circ} \\= \left(\frac{\sqrt 5 +1}{4} \right) \frac{\sqrt 3}{2} +\left(\frac{\sqrt{10-2\sqrt5}}{4}\right)\frac{1}{2} =\frac{\sqrt{10-2\sqrt 5}+\sqrt 3 +\sqrt {15}}{8}putting in original equation sin⁡3∘=12(1−10−25+3+158)=14(8−10−25−3−15)\sin3^{\circ} =\frac{1}{\sqrt 2} \left( \sqrt{1- \frac{\sqrt{10-2\sqrt 5}+\sqrt 3 +\sqrt {15}}{8} }\right) = \frac{1}{4}\left(\sqrt{8-\sqrt{10-2\sqrt 5}-\sqrt 3 -\sqrt {15} }\right) and hence cos⁡3∘=1−sin⁡23∘=148+10−25+10+3\cos 3^{\circ} = \sqrt{1-\sin^2 3^{\circ}} =\frac{1}{4}\sqrt{8+\sqrt{10-2\sqrt 5} +\sqrt{10}+\sqrt{3}} and hence we deduce easily that cot⁡9∘=cot⁡3∘(1−84+10−25+15+3)\cot9^{\circ}= \cot3^{\circ}\left(1-\frac{8}{4+\sqrt{10-2\sqrt 5}+\sqrt{15}+\sqrt{3}}\right) also cot⁡(3π20)=cot⁡(27∘=cot⁡(30−3) \cot\left(\frac{3\pi}{20}\right)=\cot(27^{\circ}=\cot(30-3) and using compound angle formula for cot(a−b)cot(a-b) we have cot⁡27∘=cot⁡3∘cot⁡30∘−1cot⁡3∘+cot⁡30∘=3cot⁡3∘−1cot⁡3∘+3\cot27^{\circ}= \frac{\cot 3^{\circ} \cot30^{\circ}-1}{\cot 3^{\circ}+\cot 30^{\circ}}=\frac{\sqrt 3 \cot3^{\circ} -1}{\cot3^{\circ}+\sqrt 3} and hence we have D=Φ(β−α,β,(α+2β+γ)−1D=\Phi(\beta-\alpha,\beta , (\alpha +2\beta +\gamma)^{-1} =2π5−1+ϕ+2log⁡(θ−8θ4+10−25+15+3)+2−110−25log⁡(3θ−1θ+3)=\frac{2\pi}{\sqrt5 -1}+\sqrt{\phi+2}\log\left(\theta -\frac{8\theta}{4+\sqrt{10-2\sqrt 5}+\sqrt{15}+\sqrt {3}}\right)+2^{-1}\sqrt{10-2\sqrt 5}\log\left(\frac{\sqrt 3 \theta -1}{\theta +\sqrt 3}\right) and thus θ=8+10−25+15+38−10−25−15−3\displaystyle \theta=\frac{\sqrt{8+\sqrt{10-2\sqrt{5}}+\sqrt{15}+\sqrt{3}}}{\sqrt{8-\sqrt{10-2\sqrt{5}}-\sqrt{15}-\sqrt{3}}} we are done

Other different approaches are highly appreciated. I wish to be corrected if any such typos or errors are spotted in the above solution. Thank you!

#Calculus

Note by Naren Bhandari
1 year, 2 months ago

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Comments

I don't know calculus but I like your solutions (both of them!)

A Former Brilliant Member - 1 year, 1 month ago

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I'm glad to know, you like it.Thank you :)

Naren Bhandari - 1 year, 1 month ago

Man I didn't understood the step where you pull out the x to the k-1 variable out from the integral.

Aruna Yumlembam - 1 year ago
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