TITLE: Why does the hard-looking integral $\int_{0}^{\infty}\frac{x\sin^2(x)}{\cosh(x)+\cos(x)}dx=1$?
QUESTION [33 upvotes]: I have to ask this question; most looking complicated definite integral yield not so nice closed form or irrational numbers or mixed of what ever ect.
Why is this particular hard looking integral gives a $1$ as an answer?
$$\int_{0}^{\infty}\frac{x\sin^2(x)}{\cosh(x)+\cos(x)}dx=1$$

REPLY [22 votes]: One may employ the generating function
$$
\frac{\sin x}{\cosh x+\cos x}=2\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\sin kx
$$
and the classic result of
$$
\int_0^\infty x^{m-1}e^{-ax} \cos bx  \ dx = \frac{\Gamma(m)}{(a^{2} + b^{2})^{m/2}}\cos\left(m\tan^{-1}\left(\frac{b}{a}\right)\right)
$$
We then have
\begin{align}
\int_{0}^{\infty}\frac{x\sin^2 x}{\cosh x+\cos x}\ dx&=2\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty}x\ e^{-kx}\sin x\sin kx\ dx\\[10pt]
&=\sum_{k=1}^\infty(-1)^{k-1}\left[\int_{0}^{\infty}x\ e^{-kx}\cos(k-1)x\ dx-\int_{0}^{\infty}x\ e^{-kx}\cos(k+1)x\ dx\right]\\[10pt]
&=\sum_{k=1}^\infty(-1)^{k-1}\left[ \frac{\cos\left(2\tan^{-1}\left(\frac{k-1}{k}\right)\right)}{k^{2}+(k-1)^2}-\frac{\cos\left(2\tan^{-1}\left(\frac{k+1}{k}\right)\right)}{k^{2}+(k+1)^2}\right]\\[10pt]
&=1
\end{align}
and the claim follows.

The latter expression is indeed a telescoping series as one may observe its partial sum equals
$$
\sum_{k=1}^n(-1)^{k-1}\left[ \frac{\cos\left(2\tan^{-1}\left(\frac{k-1}{k}\right)\right)}{k^{2}+(k-1)^2}-\frac{\cos\left(2\tan^{-1}\left(\frac{k+1}{k}\right)\right)}{k^{2}+(k+1)^2}\right]=1+\frac{\cos\left(2\tan^{-1}\left(\frac{n+1}{n}\right)\right)}{n^{2}+(n+1)^2}
$$
Notice that
$$
\cos2a+\cos2b=2\cos(a-b)\cos(a+b)
$$
and
$$
\tan^{-1}(x)+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}\qquad,\qquad x>0
$$