I have to solve this integral:
$$ \int_0^\infty \frac{x^2}{2(\cosh(x) -1)} \, dx $$
How do I convert this into a power series.
I already know that the result of that integral is $\frac{\pi^2}{3}$. That, at least to me, hints that this is a power series trick of sorts.
Thank you.
$$I=\int \frac{x^2}{\cosh(x) -1} \, dx$$
Trying integration by parts $$u=x^2 \qquad \text{and} \qquad dv=\frac{dx}{\cosh(x) -1}$$ $$du=2x\,dx\qquad \text{and} \qquad v=-\coth \left(\frac{x}{2}\right)$$ $$I=-x^2 \coth \left(\frac{x}{2}\right)+\int 2 x \coth \left(\frac{x}{2}\right)\,dx$$ Integration by parts again $$u=2x \qquad \text{and} \qquad dv=\coth \left(\frac{x}{2}\right)\,dx$$ $$du=2 dx \qquad \text{and} \qquad v=2 \log \left(\sinh \left(\frac{x}{2}\right)\right)$$ $$\int 2 x \coth \left(\frac{x}{2}\right)\,dx=4 x \log \left(\sinh \left(\frac{x}{2}\right)\right)-4\int \log \left(\sinh \left(\frac{x}{2}\right)\right)\,dx$$ For the last integral, I am stuck but Wolfram Alpha knows !
Comining all of the above plus Wolfram Alpha result, assuming $a>0$ $$J=\frac 12\int_a^\infty \frac{x^2}{\cosh(x) -1} \, dx=$$ $$\frac{1}{2} \left(4 \text{Li}_2\left(e^{-a}\right)-a \left(a+4 \log \left(1-e^{-a}\right)-a \coth \left(\frac{a}{2}\right)\right)\right)$$ and a series expansion gives $$J=\frac{\pi ^2}{3}-a+\frac{a^3}{36}+O\left(a^5\right)$$