migongoniwt

2022-06-27

I would like to evaluate the integral below
${\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{3}x}{\mathrm{cosh}x+\mathrm{cos}x}\frac{dx}{x}$

scipionhi

Expert

Since
$\frac{\mathrm{sin}\left(x\right)}{\mathrm{cosh}\left(x\right)+\mathrm{cos}\left(x\right)}=2\sum _{n\ge 1}\left(-1{\right)}^{n-1}{e}^{-nx}\mathrm{sin}\left(nx\right)$
we can combine the above with
${\mathrm{sin}}^{2}\left(x\right)\mathrm{sin}\left(nx\right)=\frac{1}{4}\left[2\mathrm{sin}\left(nx\right)+\mathrm{sin}\left(\left(2-n\right)x\right)-\mathrm{sin}\left(\left(2+n\right)x\right)\right]$ and ${\int }_{0}^{\mathrm{\infty }}{e}^{-ax}\frac{\mathrm{sin}\left(bx\right)}{x}\mathrm{d}x=\mathrm{arctan}\left(\frac{b}{a}\right)$ to get
$\begin{array}{rl}I& =\frac{1}{2}\sum _{n\ge 1}\left(-1{\right)}^{n-1}{\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-nx}}{x}\left[2\mathrm{sin}\left(nx\right)+\mathrm{sin}\left(\left(2-n\right)x\right)-\mathrm{sin}\left(\left(2+n\right)x\right)\right]\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & =\frac{1}{2}\sum _{n\ge 1}\left(-1{\right)}^{n-1}\left[\frac{\pi }{2}+\mathrm{arctan}\left(\frac{2-n}{n}\right)-\mathrm{arctan}\left(\frac{2+n}{n}\right)\right]\\ & =\frac{1}{2}\sum _{n\ge 1}\left(-1{\right)}^{n-1}\left[\mathrm{arctan}\left(n+1\right)-\mathrm{arctan}\left(n-1\right)\right]\\ & =\frac{1}{2}\left[-\mathrm{arctan}\left(0\right)+\mathrm{arctan}\left(1\right)\right]\\ & =\frac{\pi }{8}\end{array}$
as desired.

pokoljitef2

Expert

It was shown in this answer that if n is a nonnegative integer, then
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(\left(2n+1\right)x\right)}{\mathrm{cosh}\left(x\right)+\mathrm{cos}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}=\frac{\pi }{4}.$
Therefore,
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{3}\left(x\right)}{\mathrm{cosh}\left(x\right)+\mathrm{cos}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& =\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}\frac{3\mathrm{sin}\left(x\right)-\mathrm{sin}\left(3x\right)}{\mathrm{cosh}\left(x\right)+\mathrm{cos}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}\\ & =\frac{1}{4}\left(\frac{3\pi }{4}-\frac{\pi }{4}\right)\\ & =\frac{\pi }{8}.\end{array}$
In general, if m is a nonnegative integer, then
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2m+1}\left(x\right)}{\mathrm{cosh}\left(x\right)+\mathrm{cos}\left(x\right)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& \stackrel{\left(1\right)}{=}\frac{1}{{2}^{2m}}\sum _{k=0}^{m}\left(-1{\right)}^{m-k}\left(\genfrac{}{}{0}{}{2m+1}{k}\right){\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(\left(2m+1-2k\right)x\right)}{\mathrm{cosh}\left(x\right)+\mathrm{cos}\left(x\right)}\frac{\mathrm{d}x}{x}\\ & =\frac{\pi }{{2}^{2m+2}}\sum _{k=0}^{m}\left(-1{\right)}^{m-k}\left(\genfrac{}{}{0}{}{2m+1}{k}\right).\end{array}$

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