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I would like to prove that

$$\int_0^{\pi/2}\frac x {\sin x} \, \mathrm d x = 2\sum_{n\mathop = 0}^{\infty} \frac {(-1)^n}{(2n+1)^2}$$

Any hints?

user85798
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  • In general when asked to compute some integral and show it is the same as a series, it is always good to check the Taylor expansions of the integrand. The technique in the answer by Robert below is widely applicable. – A. Thomas Yerger Aug 31 '16 at 21:59
  • @AlfredYerger ...which I did, and if you look up the Taylor expansion of $x/\sin x$ you will see why that did not help me. – user85798 Aug 31 '16 at 22:42

3 Answers3

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We have that $$\int_0^{\pi/2}\frac x {\sin x} d x=\int_0^{\pi/4}\frac{2t}{2\sin t \cos t} d (2t)=2\int_0^{\pi/4}\frac{t}{\tan t \cos^2 t}d t=2\int_0^{1}\frac{\arctan s}{s}d s\\ =2\int_0^{1}\sum_{n=0}^{+\infty}\frac{(-1)^n s^{2n}}{2n+1} ds =2\sum_{n=0}^{+\infty}(-1)^n\int_0^{1}\frac{ s^{2n}}{2n+1} ds =2\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^2}.$$

Robert Z
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Since $\ds{\totald{\ln\pars{\tan\pars{x/2}}}{x} = {1 \over \sin\pars{x}}}$:

\begin{align} \color{#f00}{\int_{0}^{\pi/2}{x \over \sin\pars{x}}\,\dd x} & = -\int_{0}^{\pi/2}\ln\pars{\tan\pars{x \over 2}}\,\dd x \end{align} With the Weierstrass Tangent Half-Angle Substitution: \begin{align} \color{#f00}{\int_{0}^{\pi/2}{x \over \sin\pars{x}}\,\dd x} & = -2\int_{0}^{1}{\ln\pars{x} \over 1 + x^{2}}\,\dd x = -2\sum_{n = 0}^{\infty}\pars{-1}^{n}\ \overbrace{\int_{0}^{1}\ln\pars{x}\,x^{2n}\,\dd x} ^{\ds{-\,{1 \over \pars{2n + 1}^{2}}}} = \color{#f00}{2\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}^{2}}} \end{align}

Felix Marin
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-1

Both expressions equal $2G$, where $G$ is the Catalan constant

Paul Enta
  • 14,113