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Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $ I have stumbled upon this series: $$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}$$ The linked proof relies upon this trigamma identity. Now by rewriting the integral as: $$I=\int_0^1 \frac{\sqrt{x}\ln x}{x^2-2\cos\left(\frac{\pi}{3}\right)x+1}dx$$ And using that: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$ $$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$ $$I=\frac{1}{\sin \left(\frac{\pi}{3}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{3} (n+1) \right)\int_0^1 x^{n+1/2} \ln x dx$$ $$\text{Since} \ \int_0^1 x^p \ln x dx= -\frac{1}{(p+1)^2}$$ $$I=-\frac{2}{\sqrt 3} \sum_{n=0}^\infty \frac{\sin\left((n+1)\frac{\pi}{3}\right)}{(n+1+1/2)^2}=-\frac{8}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} $$ And well by using the previous link we can deduce that the series equals to $\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$, where $G$ is Catalan's constant.

I thought this might be a coefficient of some Fourier series, or taking the imaginary part of $\left(\sum_{n=1}^\infty \frac{e^{i\frac{n\pi}{3}}}{(2n+1)^2}\right)$, but I was not that lucky afterwards.

Is there a way to show the result without relying on that trigamma identity? Another approach to the integral would of course be enough.

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    $\begingroup$ In your definition of I, you left out the x in the denominator. $\endgroup$ Oct 20 '18 at 14:05
  • $\begingroup$ I am not sure to understand the question. You want a different approach to compute the integral? $\endgroup$
    – FDP
    Oct 21 '18 at 14:07
  • $\begingroup$ @FDP I have edited the question, hope it's clear now. $\endgroup$
    – Zacky
    Oct 21 '18 at 17:20
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    $\begingroup$ Yep and i'm pretty sure now there is an elementary way to compute the integral. It will take me several days to terminate the computation probably. $\endgroup$
    – FDP
    Oct 21 '18 at 18:27
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    $\begingroup$ I have completed the computation and i think all is correct now. It will take some hours to write down here. Thank you to show me a new formula. You'll understand my words very soon ;) $\endgroup$
    – FDP
    Oct 23 '18 at 14:59
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\begin{align*} I&=\int_0^1 \frac{\sqrt{x}\ln x} {x^2-x+1}\,dx \end{align*}

Perform the change of variable $y=\sqrt{x}$,

\begin{align*} J&=4\int_0^1 \frac{x^2\ln x} {x^4-x^2+1}\,dx\\ &=\left[\left(\frac{1}{\sqrt{3}}\ln\left(\frac{x^2-\sqrt{3}x+1}{x^2+\sqrt{3}x+1}\right)+2\arctan\left(\frac{x}{1-x^2}\right)\right)\ln x\right]_0^1-\\ &\int_0^1 \frac{1}{x\sqrt{3}}\ln\left(\frac{x^2-\sqrt{3}x+1}{x^2+\sqrt{3}x+1}\right)\,dx-2\int_0^1\frac{1}{x}\arctan\left(\frac{x}{1-x^2}\right)\,dx\\ &=-\int_0^1 \frac{1}{x\sqrt{3}}\ln\left(\frac{x^2-\sqrt{3}x+1}{x^2+\sqrt{3}x+1}\right)\,dx-2\int_0^1\frac{1}{x}\arctan\left(\frac{x}{1-x^2}\right)\,dx\\ \end{align*}

\begin{align*} L&=\int_0^1 \frac{1}{x}\ln\left(\frac{x^2-\sqrt{3}x+1}{x^2+\sqrt{3}x+1}\right)\,dx\\ &=\int_0^1 \frac{1}{x}\ln\left(\frac{\left(x^2-\sqrt{3}x+1\right)\left(x^2+\sqrt{3}x+1\right)}{\left(x^2+\sqrt{3}x+1\right)^2}\right)\,dx\\ &=\int_0^1 \frac{1}{x}\ln\left(x^4-x^2+1\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1)\right)\,dx\\ &=\int_0^1 \frac{x}{x^2}\ln\left(\frac{1+x^6}{1+x^2}\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1)\right)\,dx\\ \end{align*}

In the first integral perform the change of variable $y=x^2$,

\begin{align*}L&=\frac{1}{2}\int_0^1 \frac{1}{x}\ln\left(\frac{1+x^3}{1+x}\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ &=\frac{1}{2}\int_0^1 \frac{x^2}{x^3}\ln\left(1+x^3\right)\,dx-\frac{1}{2}\int_0^1 \frac{1}{x}\ln\left(1+x\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ \end{align*}

In the first integral perform the change of variable $y=x^3$,

\begin{align*}L&=\frac{1}{6}\int_0^1 \frac{1}{x}\ln\left(1+x\right)\,dx-\frac{1}{2}\int_0^1 \frac{1}{x}\ln\left(1+x\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ &=-\frac{1}{3}\int_0^1 \frac{1}{x}\ln\left(1+x\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ &=\frac{1}{3}\int_0^1 \frac{1}{x}\ln\left(1-x\right)\,dx-\frac{1}{3}\int_0^1 \frac{1}{x}\ln\left(1-x^2\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ \end{align*}

In the second integral perform the change of variable $y=x^2$,

\begin{align*}L&=\frac{1}{3}\int_0^1 \frac{1}{x}\ln\left(1-x\right)\,dx-\frac{1}{6}\int_0^1 \frac{1}{x}\ln\left(1-x\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ &=\frac{1}{6}\int_0^1 \frac{1}{x}\ln\left(1-x\right)\,dx-2\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\ \end{align*}

\begin{align*}M&=\int_0^1 \frac{1}{x}\ln\left(x^2+\sqrt{3}x+1\right)\,dx\\\end{align*}

Define the function $F$ on $\left[0;\sqrt{3}\right]$ by,

\begin{align*}F(a)=\int_0^1 \frac{1}{x}\ln\left(x^2+ax+1\right)\,dx\\\end{align*}

Observe that $F(\sqrt{3})=M$ and,

\begin{align*}F(0)&=\int_0^1 \frac{1}{x}\ln\left(x^2+1\right)\,dx\\ &=-\frac{1}{4}\int_0^1 \frac{1}{x}\ln\left(1-x\right)\,dx\\ \end{align*}

Since $0<a<2$,

\begin{align*}F^\prime(a)&=\int_0^1 \frac{1}{x^2+ax+1}\,dx\\ &=\left[\frac{2}{\sqrt{4-a^2}}\arctan\left(\frac{2x+a}{\sqrt{4-a^2}}\right)\right]_0^1\\ &=\frac{2}{\sqrt{4-a^2}}\arctan\left(\frac{2+a}{\sqrt{4-a^2}}\right)-\frac{2}{\sqrt{4-a^2}}\arctan\left(\frac{a}{\sqrt{4-a^2}}\right)\\ &=\frac{2}{\sqrt{4-a^2}}\arctan\left(\sqrt{\frac{2-a}{2+a}}\right)\\ \end{align*}

\begin{align*}F(\sqrt{3})-F(0)=\int_0^{\sqrt{3}}\frac{2}{\sqrt{4-a^2}}\arctan\left(\sqrt{\frac{2-a}{2+a}}\right)\,da\\ \end{align*}

Perform the change of variable $y=\sqrt{\frac{2-a}{2+a}}$,

\begin{align*}F(\sqrt{3})-F(0)&=4\int_{2-\sqrt{3}}^1 \frac{\arctan y}{1+y^2}\,dy\\ &=2\Big[\arctan^2(x)\Big]_{2-\sqrt{3}}^1\\ &=2\times \frac{\pi^2}{4^2} -2\times \frac{\pi^2}{12^2}\\ &=\frac{\pi^2}{9} \end{align*}

Since,

\begin{align*}\int_0^1 \frac{1}{x}\ln\left(1-x\right)\,dx=-\frac{\pi^2}{6}\end{align*}

Then, \begin{align*}M&=F(\sqrt{3})\\ &=F(0)+\frac{\pi^2}{9}\\ &=-\frac{1}{4}\times -\frac{\pi^2}{6}+\frac{\pi^2}{9}\\ &=\frac{11\pi^2}{72} \end{align*}

Therefore,

\begin{align*}L&=-\frac{1}{6}\times -\frac{\pi^2}{6}-2M\\ &=\frac{1}{6}\times -\frac{\pi^2}{6}-2\times \frac{11\pi^2}{72}\\ &=-\frac{\pi^2}{3} \end{align*}

\begin{align*} K&=\int_0^1 \frac{\arctan\left(\frac{x}{1-x^2}\right)}{x}\,dx\\ \end{align*}

Perform the change of variable $x=\tan\left(\frac{t}{2}\right) $,

\begin{align*} K&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(\frac{1}{2}\tan t\right)}{\sin t}\,dt \end{align*}

Define the function $H$ on $\left[\frac{1}{2};1\right]$ by,

\begin{align*}H(a)&=\int_0^{\frac{\pi}{2}} \frac{\arctan\left(a\tan t\right)}{\sin t}\,dt\end{align*}

Observe that $K=H\left(\dfrac{1}{2}\right)$ and,

\begin{align*}H(1)&=\int_0^{\frac{\pi}{2}} \frac{t}{\sin t}\,dt\\ &=\Big[t\ln\left(\tan\left(\frac{t}{2} \right)\right)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\ &=-\int_0^{\frac{\pi}{2}}\ln\left(\tan\left(\frac{t}{2} \right)\right)\,dt\\ \\\end{align*}

Perform the change of variable $x=\dfrac{t}{2}$,

\begin{align*}H(1)&=-2\int_0^{\frac{\pi}{4}}\ln\left(\tan\left(t \right)\right)\,dt\\ &=2\text{G} \\\end{align*}

\begin{align*}H^\prime (a)&=\int_0^{\frac{\pi}{2}} \frac{\cos x}{1-(1-a^2)\sin^2 x}\,dt\\ &=\left[\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sin(x)\sqrt{1-a^2}}{1-\sin(x)\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}}\\ &=\frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right) \end{align*}

Therefore,

\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=\int_{\frac{1}{2}}^1 \frac{1}{2\sqrt{1-a^2}}\ln\left(\frac{1+\sqrt{1-a^2}}{1-\sqrt{1-a^2}}\right)\,da\end{align*}

Perform the change of variable $y=\arctan\left(\sqrt{\dfrac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)$

\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=-2\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy\\ &=-2\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy+\int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy \end{align*}

But, it's well known that:

\begin{align*}\int_0^{\frac{\pi}{4}} \ln\left(\tan y\right)\,dy=-\text{G}\\ \int_0^{\frac{\pi}{12}} \ln\left(\tan y\right)\,dy=-\frac{2}{3}\text{G}\\ \end{align*}

(see Integral: $\int_0^{\pi/12} \ln(\tan x)\,dx$ )

Therefore,

\begin{align*}H(1)-H\left(\frac{1}{2}\right)&=2\text{G}+2\times -\frac{2}{3}\text{G}\\ &=\frac{2}{3}\text{G} \end{align*}

Thus,

\begin{align*} K&=H\left(\frac{1}{2}\right)\\ &=H(1)-\frac{2}{3}\text{G}\\ &=2\text{G}-\frac{2}{3}\text{G}\\ &=\frac{4}{3}\text{G} \end{align*}

\begin{align*}I&=-\frac{1}{\sqrt{3}}L-2K\\ &=-\frac{1}{\sqrt{3}}\times -\frac{\pi^2}{3}-2\times \frac{4}{3}\text{G}\\ &=\boxed{\frac{\sqrt{3}\pi^2}{9}-\frac{8}{3}\text{G}} \end{align*}

PS:

Actually,

\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\int_0^1 \frac{\arctan \left(x^3\right)}{x}\,dx\end{align}

In the latter integral perform the change of variable $y=x^3$,

\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx-\int_0^1 \frac{\arctan x}{x}\,dx=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx\end{align}

therefore,

\begin{align}\int_0^1 \frac{\arctan\left( \frac{x}{1-x^2}\right)}{x}\,dx&=\frac{1}{3}\int_0^1 \frac{\arctan x}{x}\,dx+\int_0^1 \frac{\arctan x}{x}\,dx\\ &=\frac{4}{3}\int_0^1 \frac{\arctan x}{x}\,dx\\ &=\frac{4}{3}\text{G} \end{align}

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This is a major revision of the last few lines. The conclusion is that

$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} =\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144} $

where $\psi^{(1)}$ is a polygamma function (reference below).

$\begin{array}\\ \sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} &=\sum_{n=1}^\infty \left(\frac{\sin\left((3n-2)\frac{\pi}{3}\right)}{(2(3n-2)+1)^2}+\frac{\sin\left((3n-1)\frac{\pi}{3}\right)}{(2(3n-1)+1)^2}+\frac{\sin\left(3n\frac{\pi}{3}\right)}{(2(3n)+1)^2}\right)\\ &=\sum_{n=1}^\infty \left(\frac{\sin\left(-2\frac{\pi}{3}\right)}{(6n-3)^2}+\frac{\sin\left(-\frac{\pi}{3}\right)}{(6n-1)^2}+\frac{\sin\left(n\pi\right)}{(6n+1)^2}\right)\\ &=\sum_{n=1}^\infty \left(-\frac{\sqrt{3}/2}{(6n-3)^2}+\frac{\sqrt{3}/2}{(6n-1)^2}\right)\\ &=\frac{\sqrt{3}}{2}\sum_{n=1}^\infty \left(-\frac1{(6n-3)^2}+\frac1{(6n-1)^2}\right)\\ &=\frac{\sqrt{3}}{2}\sum_{n=1}^\infty \left(\frac1{(6n-1)^2}-\frac1{(6n-3)^2}\right)\\ &=\frac{\sqrt{3}}{2}\left(\sum_{n=1}^\infty \frac1{(6n-1)^2}-\frac19\sum_{n=1}^\infty\frac1{(2n-1)^2}\right)\\ &=\frac{\sqrt{3}}{2}\left(\frac1{36}\sum_{n=1}^\infty \frac1{(n-1/6)^2}-\frac19(1-\frac14)\zeta(2)\right)\\ &=\frac{\sqrt{3}}{2}\left(\frac1{36}\sum_{n=0}^\infty \frac1{(n+5/6)^2}-\frac19\frac34\frac{\pi^2}{6}\right)\\ &=\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144}\\ \end{array} $

$\psi^{(1)}(z)$ is a polygamma function: https://en.wikipedia.org/wiki/Polygamma_function

$\psi^{(m)}(z) =(-1)^{m+1}m!\sum_{n=0}^{\infty} \dfrac1{(z+n)^{m+1}} $ so that $\psi^{(1)}(z) =\sum_{n=0}^{\infty} \dfrac1{(z+n)^2} $

Also

$\begin{array}\\ \zeta(m) &=\sum_{n=1}^{\infty} \dfrac1{n^m}\\ &=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\sum_{n=1}^{\infty} \dfrac1{(2n)^m}\\ &=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\dfrac1{2^m}\sum_{n=1}^{\infty} \dfrac1{n^m}\\ &=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\dfrac1{2^m}\zeta(m)\\ \end{array} $

so

$\zeta(m)(1-2^{-m}) =\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m} $.

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    $\begingroup$ Thanks for this, but there is a mistake because: $$\sin \left(n\pi -\frac{2\pi}{3}\right)=\sin \left(n\pi -\frac{\pi}{3}\right)=(-1)^{n+1}\frac{\sqrt 3}{2}$$ Thus we have to evaluate: $$\frac{\sqrt 3}{2} \sum_{n=1}^\infty (-1)^{n+1}\left(\frac{1}{(6n-3)^2} +\frac{1}{(6n-1)^2}\right)$$ Well the first sum is trivial using the definition of the Catalan constant and it's equal to $\frac{G}{9}$. And we are left with: $$\frac{\sqrt 3G}{18} + \sum_{n=1}^\infty (-1)^{n+1}\left(\frac{1}{(6n-1)^2} \right)$$ And unfortunately the last sum is equal to $\frac{1}{144}$ times the linked trigamma identity :( $\endgroup$
    – Zacky
    Oct 22 '18 at 20:36
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    $\begingroup$ Thanks for catching my mistake. $\endgroup$ Oct 22 '18 at 22:22
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Integrate by parts \begin{align} \int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =& \int_0^1 \ln x \>d\left( \tan^{-1} \frac{\sqrt x} {1-x} -\frac1{\sqrt3 } \tanh^{-1} \frac{\sqrt {3x}} {1+x} \right)\\ =&\frac1{\sqrt3 }I_1 - I_2\tag1 \end{align} where \begin{align} I_1&=\int_0^1 \frac{\tanh^{-1} \frac{\sqrt {3x}} {1+x}}x dx\\ I_2 & = \int_0^1 \frac{\tan^{-1} \frac{\sqrt {x}} {1-x}}x dx = \underset{\sqrt x\to x}{\int_0^1 \frac{\tan^{-1} \sqrt x}x dx } + \underset{\sqrt {x^3}\to x}{\int_0^1 \frac{\tan^{-1} \sqrt {x^3}}x }dx \\ &= \left(2+ \frac23 \right) \int_0^1 \frac{\tan^{-1} x}x dx = \frac83G\tag2 \end{align} Evaluate $I_1$ with $J(a) = \int_0^1 \frac{\tanh^{-1} \frac{2a\sqrt {x}} {1+x}}x dx$ $$J’(a) =\int_0^1 \frac{2(\frac1{\sqrt x}+\sqrt x)dx}{(x+1)^2-(2a\sqrt x)^2} =\frac{2 \tan^{-1}\frac{\sqrt{(1-a^2)x}}{1-x}\bigg|_0^1} {\sqrt{1-a^2}}=\frac\pi{\sqrt{1-a^2}} $$ Then $$I_1 = J(\frac{\sqrt3}2)=\int_0^{ \frac{\sqrt3}2}J’(a)da =\int_0^{ \frac{\sqrt3}2} \frac\pi{\sqrt{1-a^2}} da =\frac{\pi^2}3\tag3 $$ Plug (2) and (3) into (1) to obtain $$\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx = \frac{\pi^2}{3\sqrt3}-\frac83G$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}} = {G \over \root{3}} - {\pi^{2} \over 24}} \approx 0.1176:\ {\Large ?}}$. $\ds{G = 0.9159\ldots}$ is the Catalan Constant.


\begin{align} &\bbox[5px,#ffd]{\sum_{n = 1}^{\infty} {\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}}} = \sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}} + \sum_{n = 6}^{11}{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}} + \sum_{n = 12}^{17}{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}} + \cdots \\[5mm] = &\ \sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}} + \sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over \bracks{2n + 2\pars{6} + 1}^{\, 2}} + \sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over \bracks{2n + 4\pars{6} + 1}^{\, 2}} + \cdots \\[5mm] = &\ \sum_{n = 0}^{5}\sin\pars{n\,{\pi \over 3}} \sum_{k = 0}^{\infty}{1 \over \pars{2n + 12k + 1}^{2}} = {1 \over 144}\sum_{n = 0}^{5}\sin\pars{n\,{\pi \over 3}}\, \Psi\, '\pars{2n + 1 \over 12} \\[5mm] = &\ {G \over \root{3}}\ -\ \underbrace{\bracks{80G + \Psi\, '\pars{11 \over 12} - \Psi\, '\pars{5 \over 12}}{\root{3} \over 288}} _{\ds{=\ {\pi^{2} \over 24}}} \end{align} I'll $\ds{\underline{\mbox{have}}}$ to prove that $\ds{\color{red}{\Psi\, '\pars{11 \over 12} - \Psi\, '\pars{5 \over 12} \color{black}{=} 4\root{3}\pi^{2} - 80G}}$
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