# Show $\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1)$

I am trying to prove this interesting integral $$I:=\int_0^\infty \log(1+x^2) \frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx=4\sqrt 2-\frac{16}{\pi}+\frac{8\sqrt 2}{\pi}\log(\sqrt 2+1).$$ I tried to write $$\int_0^\infty \log(1+x)\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx+\int_0^\infty \log(1-x)\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx$$ and now using $$\int_0^\infty \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx-\int_0^\infty \sum_{n=1}^\infty \frac{x^n}{n}\frac{\cosh \frac{\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx$$ and now introducing a parameter $$I(a)=\int_0^\infty \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n\frac{\cosh \frac{a \pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx-\int_0^\infty \sum_{n=1}^\infty \frac{x^n}{n}\frac{\cosh \frac{a\pi x}{4}}{\sinh^2 \frac{\pi x}{4}}dx.$$ But writing $I'(a)$ didn't simplify much. The substitution $y=\sinh \pi x/4$ also was of no use because of the $x^n$ factor. So How can we prove this interesting integral? Thanks

• Notice that $\cosh t=\sinh't$, and $\bigg(\dfrac1u\bigg)'=-\dfrac1{u^2}$ – Lucian May 20 '14 at 20:48
• Is there a typo in the original question or an error in the calculation, $1+x^2$ does not factor. – Rene Schipperus May 20 '14 at 21:02
• – Rene Schipperus May 20 '14 at 21:05
• @ Rene thanks I posted that question though:) – Jeff Faraci May 20 '14 at 21:05
• @Rene no typo I made an error trying to solve that's all . Which is why I am looking for a solution. Thanks for noticing the mistake – Jeff Faraci May 20 '14 at 21:06

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I\equiv\int_{0}^{\infty}\ln\pars{1 + x^{2}}\, {\cosh\pars{\pi x/4} \over \sinh^{2}\pars{\pi x/4}}\,\dd x =4\root{2} - {16 \over \pi} + {8\root{2} \over \pi}\,\ln\pars{\root{2} + 1}: \ {\large ?}}$
Zeros $\ds{\braces{x_{n}}}$ of $\ds{\sinh\pars{\pi x \over 4}}$ are given by $\quad\ds{x_{n} = 4n\ic\,,\quad n \in {\mathbb Z}\quad}$ such that \begin{align} I&={4 \over \pi}\bracks{2\pi\ic\,{\ic \over \pars{\ic + \ic}\sinh\pars{\pi\ic/4}} +\sum_{n = 1}^{\infty}2\pi\ic\,{4n\ic\over \pars{4n\ic - \ic}\pars{4n\ic + \ic}\cosh\pars{\pi\bracks{4n\ic}/4}\pars{\pi/4}}} \\[3mm]&=4\root{2} +{8 \over \pi}\sum_{n = 1}^{\infty}\pars{-1}^{n}\, {n \over \pars{n - 1/4}\pars{n + 1/4}} =4\root{2} +{4 \over \pi}\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n - 1/4}} +{4 \over \pi}\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + 1/4}} \end{align}
With $\ds{a \not\in {\mathbb Z}}$: \begin{align} &\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + a}} =\sum_{n = 0}^{\infty}\pars{{1 \over 2n + 2 + a} - {1 \over 2n + 1 + a}} =-\sum_{n = 0}^{\infty}{1 \over \pars{2n + 2 + a}\pars{2n + 1 + a}} \\[3mm]&=-\,{1 \over 4}\sum_{n = 0}^{\infty} {1 \over \pars{n + 1 + a/2}\pars{n + 1/2 + a/2}} =-\,{1 \over 4}\,{\Psi\pars{1 + a/2} - \Psi\pars{1/2 + a/2} \over \pars{1 + a/2} - \pars{1/2 + a/2}} \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function $\bf\mbox{6.3.1}$.
$$\color{#00f}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n + a}} =\half\,\bracks{\Psi\pars{\half + {a \over 2}} - \Psi\pars{1 + {a \over 2}}}$$