TITLE: Prove $\pi^2\int_0^\infty\frac{x\sin^4\pi x}{\cos\pi x+\cosh\pi x}dx=e^2\int_0^\infty\frac{x\sin^4ex}{\cos ex+\cosh ex}dx=\frac{176}{225}$
QUESTION [16 upvotes]: Marco Cantarini and Jack D'Aurizio proved hard-looking integrals (see Marco and Jack) in my recent two posts.  
This is our final hard-looking integral that yield a rational answer:

$$\pi^2\int_{0}^{\infty}\frac{x\sin^4(x\pi)}{\cos(x\pi)+\cosh(x\pi)}dx=e^2\int_{0}^{\infty}\frac{x\sin^4(xe)}{\cos(xe)+\cosh(xe)}dx=\frac{176}{225}\tag1$$

Can anyone provide us a prove of $(1)$?

REPLY [7 votes]: Taking the cue from Sophie Agnesi and doing a similar manner as my previous answer in your post, then
\begin{align}
S_A&=\int_{0}^{\infty}\frac{x\sin^4 x}{\cosh x+\cos x}\ dx\\[10pt]
&=2\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty}x\ e^{-kx}\sin^3 x\sin kx\ dx\\[10pt]
&=\frac{1}{2}\sum_{k=1}^\infty(-1)^{k-1}\left[3\int_{0}^{\infty}x\ e^{-kx}\sin x\sin kx\ dx-\int_{0}^{\infty}x\ e^{-kx}\sin3 x\sin kx\ dx\right]\\[10pt]
&=\frac{3}{4}-\frac{1}{4}\sum_{k=1}^\infty(-1)^{k-1}\left[\int_{0}^{\infty}x\ e^{-kx}\cos(k-3)x\ dx-\int_{0}^{\infty}x\ e^{-kx}\cos(k+3)x\ dx\right]\\[10pt]
&=\frac{3}{4}-\frac{1}{4}\sum_{k=1}^\infty(-1)^{k-1}\left[ \frac{\cos\left(2\tan^{-1}\left(\frac{k-3}{k}\right)\right)}{k^{2}+(k-3)^2}-\frac{\cos\left(2\tan^{-1}\left(\frac{k+3}{k}\right)\right)}{k^{2}+(k+3)^2}\right]\\[10pt]
&=\frac{3}{4}-\frac{1}{4}\left(-\frac{29}{225}\right)\\[10pt]
&=\frac{176}{225}
\end{align}
and the claim follows. I leave the latter sum to you as a brainstorming.