A closed-form of $\frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx$

Question

I am looking for a closed-form of this integral

\begin{equation} \frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx \end{equation}

I can rewrite the integral into \begin{equation} \int_0^\infty\frac{x^2(e^{2x}\cos x-1)}{e^{4x}-2e^{2x}\cos x+1}\,dx \end{equation} But I am stuck for the next step. I have a strong feeling the integral involving gamma or beta function but I am unable to prove it. Could anyone here please help me? Any help would be greatly appreciated. Thank you.

Answer 1

Edit: The previous answer was wrong because it began with the incorrect series expansion.

The correct trigonometric series is the following:

$$\frac{1}{(p^2-1)}+\frac{p^2+1}{(p^2-1)}\sum_{k=1}^{\infty}\frac{\cos{kx}}{p^k}=\frac{p\cos{x}}{p^2-2p\cos{x}+1};~where~p>1,$$

and

$$\frac{1}{p^2-1}+\frac{2}{p^2-1}\sum_{k=1}^{\infty}\frac{\cos{kx}}{p^k}=\frac{1}{p^2-2p\cos{x}+1};~where~p>1.$$

Subtracting the two gives,

$$\sum_{k=1}^{\infty}\frac{\cos{kx}}{p^k}=\frac{p\cos{x}-1}{p^2-2p\cos{x}+1};~where~p>1.$$

With $p=e^{2x}$,

$$\begin{align} \int_{0}^{\infty}\mathrm{d}x\,\frac{x^2(e^{2x}\cos{x}-1)}{e^{4x}-2e^{2x}\cos{x}+1} &=\int_{0}^{\infty}\mathrm{d}x\,\sum_{k=1}^{\infty}x^2e^{-2kx}\cos{kx}\\ &=\sum_{k=1}^{\infty}\int_{0}^{\infty}\mathrm{d}x\,x^2e^{-2kx}\cos{kx}\\ &=\sum_{k=1}^{\infty}\left[\frac{\partial^2}{\partial k^2}\int_{0}^{\infty}\mathrm{d}x\,\frac14e^{-2kx}\cos{nx}\right]_{n=k}\\ &=\sum_{k=1}^{\infty}\left[\frac{\partial^2}{\partial k^2}\frac14\cdot\frac{2k}{4k^2+n^2}\right]_{n=k}\\ &=\sum_{k=1}^{\infty}\left[\frac{4k(4k^2-3n^2)}{(4k^2+n^2)^3}\right]_{n=k}\\ &=\sum_{k=1}^{\infty}\frac{4}{125k^3}\\ &=\frac{4}{125}\zeta{(3)}. \end{align}$$

Thanks to Tunk-Fey for pointing out the erroneous value I had beforehand.