Ernest Ryland

2022-01-01

Sovle integral:
${\int }_{0}^{\mathrm{\infty }}\mathrm{log}\left(1+{x}^{2}\right)\frac{\text{cosh}\frac{\pi x}{2}}{{\text{sinh}}^{2}\frac{\pi x}{2}}dx=2-\frac{4}{\pi }$

chumants6g

Integrating by parts shows that the integral is equivalent to showing that
${\int }_{0}^{\mathrm{\infty }}\frac{x}{1+{x}^{2}}\frac{1}{\text{sinh}\frac{\pi x}{2}}dx=\frac{\pi }{2}-1$
Let $f\left(z\right)=\frac{z}{1+{z}^{2}}\frac{1}{\text{sinh}\frac{\pi z}{2}}$ and integrate around a rectangle with vertices at $±N$ and $±N+i\left(2N+1\right)$ where N is some positive integer.
As N goes to infinity through the integers, the integral vanishes on the left and right sides of the rectangle and along the top of the rectangle.
In particular, the absolute value of the integral along the top of the rectangle is bounded by
$\frac{3N+1}{{\left(2N+1\right)}^{2}-1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\text{cosh}\frac{\pi x}{2}}dx=\frac{6N+2}{{\left(2N+1\right)}^{2}-1}\to 0$ as N
Then
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x}{1+{x}^{2}}\frac{1}{\text{sinh}\frac{\pi x}{2}}dx=2\pi i\left(Res\left[f\left(z\right),i\right]+\sum _{k=1}^{\mathrm{\infty }}Res\left[f\left(z\right),2ki\right]\right)$
where
$Res\left[f\left(z\right),i\right]=\underset{z\to i}{lim}\frac{z}{z+i}\frac{1}{\text{sinh}\frac{\pi x}{2}}=\frac{1}{2i}$
and
$Res\left[f\left(z\right),2ki\right]=\underset{z\to 2ki}{lim}\frac{z}{2z\text{sinh}\frac{\pi z}{2}+\left(1+{z}^{2}\right)\frac{\pi }{2}\text{cosh}\frac{\pi z}{2}}$
$=\frac{4i}{\pi }\frac{{\left(-1\right)}^{k}k}{1-4{k}^{2}}$
And notice that
$\sum _{k=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{k}k}{1-4{k}^{2}}=-\frac{1}{4}\sum _{k=1}^{\mathrm{\infty }}\left(\frac{{\left(-1\right)}^{k}}{2k+1}+\frac{{\left(-1\right)}^{k}}{2k-1}\right)$

Daniel Cormack

I will solve the general form
${\int }_{0}^{\mathrm{\infty }}\frac{x}{{x}^{2}+1}\frac{dx}{\text{sinh}\left(ax\right)}={\int }_{0}^{\mathrm{\infty }}\frac{x}{{x}^{2}+{a}^{2}}\frac{dx}{\text{sinh}\left(x\right)}$
$={\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}{e}^{-at}\frac{\mathrm{sin}\left(xt\right)}{\text{sinh}\left(x\right)}dtdx$
$={\int }_{0}^{\mathrm{\infty }}{e}^{-at}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(xt\right)}{\text{sinh}\left(x\right)}dxdt$

$={\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-zx}\left(1-{e}^{-2x}\right)}{{e}^{-2x}+1}dx$
By spliting the integral we have
${\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-zx}}{{e}^{-2x}+1}dx=\sum _{n\ge 0}{\int }_{0}^{\mathrm{\infty }}{e}^{-x\left(2n+z\right)}dx$
$=\sum _{n\ge 0}\frac{{\left(-1\right)}^{n}}{2n+z}$
$=\frac{1}{4}\left(\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\psi \left(1+\frac{z}{4}\right)-\psi \left(\frac{z}{4}\right)\right)$
$=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{z}{4}\right)-\frac{1}{2}\psi \left(\frac{z}{4}\right)-z$
$=\frac{1}{2}\psi \left(\frac{1}{2}+\frac{a}{2\pi }\right)-\frac{1}{2}\psi \left(\frac{a}{2}$

Vasquez

$\begin{array}{}I\equiv {\int }_{0}^{\mathrm{\infty }}\mathrm{ln}\left(1+{x}^{2}\right)\frac{\mathrm{cosh}\left(\pi x/2\right)}{{\mathrm{sinh}}^{2}\left(\pi x/2\right)}dx=2-\frac{4}{\pi }\\ I=-\frac{1}{\pi }{\int }_{x\to -\mathrm{\infty }}^{x\to \mathrm{\infty }}\mathrm{ln}\left(1+{x}^{2}\right)d\left[\frac{1}{\mathrm{sinh}\left(\pi x/2\right)}\right]=\frac{2}{\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\mathrm{sinh}\left(\pi x/2\right)}\frac{x}{{x}^{2}+1}dx\\ =\frac{2}{\pi }\left[2\pi i\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\pi \mathrm{cosh}\left(n\pi i\right)/2}\frac{2ni}{\left(2ni{\right)}^{2}+1}+2\pi i\frac{1}{\mathrm{sinh}\left(\pi i/2\right)}\frac{i}{2i}\right]\\ =\frac{16}{\pi }\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{n}{4{n}^{2}-1}+2\\ =2+\frac{4}{\pi }\left[\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n-1}+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+1}\right]\\ =2+\frac{4}{\pi }\left[-\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+1}\right]\\ =2+\frac{4}{\pi }\left[-1-\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{2n+1}\right]\\ =2-\frac{4}{\pi }\\ \approx 0.7268\end{array}$