TITLE: Abel's Integral $\int_{0}^{\infty} \frac{x}{\sinh(\pi x)(1+x^2)} \,dx$
QUESTION [8 upvotes]: According to Wolfram Alpha, "Abel's Integral" is the following, and it's value is found in the method indicated below. I am unable to see how line 6 is obtained. $$$$

REPLY [2 votes]: You can use Parseval's theorem
$$ \int_{-\infty}^{\infty}f(x)\bar{g}(x)dx=\int_{-\infty}^{\infty}\hat{f}(s)\hat{\bar{g}}(s)ds $$
to evaluate this integral. Let 
$$ f(x)=\frac{x}{\sinh(\pi x)}, g(x)=\frac{1}{1+x^2}. $$
Clearly
$$  \hat{g}(s)=\sqrt{\frac{\pi}{2}}e^{-|s|}. $$
From this, one has
$$ \hat{f}(s)=\frac{1}{2\sqrt{2\pi}\cosh^2(\frac{s}{2})}. $$
So
\begin{eqnarray}
\int_{0}^{\infty}f(x)g(x)dx&=&\frac{1}{2}\int_{-\infty}^{\infty}f(x)\bar{g}(x)dx\\
&=&\frac{1}{2}\int_{-\infty}^{\infty}\hat{f}(s)\hat{\bar{g}}(s)ds\\
&=&\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{2\sqrt{2\pi}\cosh^2(\frac{s}{2})}\sqrt{\frac{\pi}{2}}e^{-|s|}ds\\
&=&\frac12(2\ln2-1).
\end{eqnarray}