migongoniwt

Answered

2022-06-27

I would like to evaluate the integral below

${\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{3}x}{\mathrm{cosh}x+\mathrm{cos}x}\frac{dx}{x}$

${\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{3}x}{\mathrm{cosh}x+\mathrm{cos}x}\frac{dx}{x}$

Answer & Explanation

scipionhi

Expert

2022-06-28Added 25 answers

Since

$\frac{\mathrm{sin}(x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}=2\sum _{n\ge 1}(-1{)}^{n-1}{e}^{-nx}\mathrm{sin}(nx)$

we can combine the above with

${\mathrm{sin}}^{2}(x)\mathrm{sin}(nx)=\frac{1}{4}[2\mathrm{sin}(nx)+\mathrm{sin}((2-n)x)-\mathrm{sin}((2+n)x)]$ and ${\int}_{0}^{\mathrm{\infty}}{e}^{-ax}\frac{\mathrm{sin}(bx)}{x}\mathrm{d}x=\mathrm{arctan}\left(\frac{b}{a}\right)$ to get

$\begin{array}{rl}I& =\frac{1}{2}\sum _{n\ge 1}(-1{)}^{n-1}{\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-nx}}{x}[2\mathrm{sin}(nx)+\mathrm{sin}((2-n)x)-\mathrm{sin}((2+n)x)]\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & =\frac{1}{2}\sum _{n\ge 1}(-1{)}^{n-1}[\frac{\pi}{2}+\mathrm{arctan}\left(\frac{2-n}{n}\right)-\mathrm{arctan}\left(\frac{2+n}{n}\right)]\\ & =\frac{1}{2}\sum _{n\ge 1}(-1{)}^{n-1}[\mathrm{arctan}(n+1)-\mathrm{arctan}(n-1)]\\ & =\frac{1}{2}[-\mathrm{arctan}(0)+\mathrm{arctan}(1)]\\ & =\frac{\pi}{8}\end{array}$

as desired.

$\frac{\mathrm{sin}(x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}=2\sum _{n\ge 1}(-1{)}^{n-1}{e}^{-nx}\mathrm{sin}(nx)$

we can combine the above with

${\mathrm{sin}}^{2}(x)\mathrm{sin}(nx)=\frac{1}{4}[2\mathrm{sin}(nx)+\mathrm{sin}((2-n)x)-\mathrm{sin}((2+n)x)]$ and ${\int}_{0}^{\mathrm{\infty}}{e}^{-ax}\frac{\mathrm{sin}(bx)}{x}\mathrm{d}x=\mathrm{arctan}\left(\frac{b}{a}\right)$ to get

$\begin{array}{rl}I& =\frac{1}{2}\sum _{n\ge 1}(-1{)}^{n-1}{\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-nx}}{x}[2\mathrm{sin}(nx)+\mathrm{sin}((2-n)x)-\mathrm{sin}((2+n)x)]\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\\ & =\frac{1}{2}\sum _{n\ge 1}(-1{)}^{n-1}[\frac{\pi}{2}+\mathrm{arctan}\left(\frac{2-n}{n}\right)-\mathrm{arctan}\left(\frac{2+n}{n}\right)]\\ & =\frac{1}{2}\sum _{n\ge 1}(-1{)}^{n-1}[\mathrm{arctan}(n+1)-\mathrm{arctan}(n-1)]\\ & =\frac{1}{2}[-\mathrm{arctan}(0)+\mathrm{arctan}(1)]\\ & =\frac{\pi}{8}\end{array}$

as desired.

pokoljitef2

Expert

2022-06-29Added 9 answers

It was shown in this answer that if n is a nonnegative integer, then

${\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}((2n+1)x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}=\frac{\pi}{4}.$

Therefore,

$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{3}(x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& =\frac{1}{4}{\int}_{0}^{\mathrm{\infty}}\frac{3\mathrm{sin}(x)-\mathrm{sin}(3x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}\\ & =\frac{1}{4}(\frac{3\pi}{4}-\frac{\pi}{4})\\ & =\frac{\pi}{8}.\end{array}$

In general, if m is a nonnegative integer, then

$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2m+1}(x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& \stackrel{(1)}{=}\frac{1}{{2}^{2m}}\sum _{k=0}^{m}(-1{)}^{m-k}{\textstyle (}\genfrac{}{}{0ex}{}{2m+1}{k}{\textstyle )}{\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}((2m+1-2k)x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\frac{\mathrm{d}x}{x}\\ & =\frac{\pi}{{2}^{2m+2}}\sum _{k=0}^{m}(-1{)}^{m-k}{\textstyle (}\genfrac{}{}{0ex}{}{2m+1}{k}{\textstyle )}.\end{array}$

${\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}((2n+1)x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}=\frac{\pi}{4}.$

Therefore,

$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{3}(x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& =\frac{1}{4}{\int}_{0}^{\mathrm{\infty}}\frac{3\mathrm{sin}(x)-\mathrm{sin}(3x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}\\ & =\frac{1}{4}(\frac{3\pi}{4}-\frac{\pi}{4})\\ & =\frac{\pi}{8}.\end{array}$

In general, if m is a nonnegative integer, then

$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{sin}}^{2m+1}(x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{d}x}{x}& \stackrel{(1)}{=}\frac{1}{{2}^{2m}}\sum _{k=0}^{m}(-1{)}^{m-k}{\textstyle (}\genfrac{}{}{0ex}{}{2m+1}{k}{\textstyle )}{\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{sin}((2m+1-2k)x)}{\mathrm{cosh}(x)+\mathrm{cos}(x)}\frac{\mathrm{d}x}{x}\\ & =\frac{\pi}{{2}^{2m+2}}\sum _{k=0}^{m}(-1{)}^{m-k}{\textstyle (}\genfrac{}{}{0ex}{}{2m+1}{k}{\textstyle )}.\end{array}$

Most Popular Questions