Abel's Integral $\int_{0}^{\infty} \frac{x}{\sinh(\pi x)(1+x^2)} \,dx$ 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. $$$$ 
 A: Note that $$\frac{b}{{{b^2} + 1}} = \int_0^\infty  {{e^{ - bx}}\cos xdx} $$
Hence
$$\int_0^\infty  {\frac{x}{{{x^2} + 1}}{e^{ - ax}}dx}  = \int_0^\infty  {\int_0^\infty  {{e^{ - xt}}(\cos t){e^{ - ax}}dt} dx}  = \int_0^\infty  {\frac{{\cos t}}{{t + a}}dt}$$
this gives
$$\text{Ci}(\pi n) = {( - 1)^{n + 1}}\int_0^\infty  {\frac{x}{{{x^2} + 1}}{e^{ - \pi nx}}dx}$$
Summing over $n=1,3,5...$ gives 
$$\sum_{n = 1,3,5...}\text{Ci}(\pi n) = \int_{0}^{\infty} \frac{x}{(x^2+1)(e^{\pi x}-e^{-\pi x})} dx$$
This is line 6.

For the evaluation of the integral, consider
$$ \mathcal{I} = \int_{-\infty}^{\infty} \frac{x}{\sinh(\pi x)(1+x^2)}dx $$
Consider the rectangular contour with vertices $R, R+Ri, -R+Ri, -R$, where $R=2N+1/2$ with $N$ a very large integer. When $R$ tends to infinity, the following three integrals tend to 0
$$\int_{R}^{R+Ri} \frac{x}{\sinh(\pi x)(1+x^2)}dx \to 0$$
$$\int_{R+Ri}^{-R+Ri} \frac{x}{\sinh(\pi x)(1+x^2)}dx \to 0$$
$$\int_{-R+Ri}^{-R} \frac{x}{\sinh(\pi x)(1+x^2)}dx \to 0$$
Note that the function $\dfrac{z}{\sinh(\pi z)(1+z^2)}$ has simple pole at $z=ni$, with $n\geq 2$ and a double pole at $z=i$, the residue at $z=ni$ being $\dfrac{(-1)^n(ni)}{\pi(1-n^2)}$, and at $z=i$ being $\frac{i}{4\pi}$.
Hence $$\begin{aligned}
\mathcal{I} &= 2\pi i\frac{i}{4\pi} + 2\pi i\sum_{n = 2}^{\infty}\frac{(-1)^n(ni)}{\pi(1-n^2)} \\
&= -\frac{1}{2}+2\sum_{n = 2}^{\infty}\frac{(-1)^nn}{n^2-1} \\
&= -\frac{1}{2}+\sum_{n = 2}^{\infty}\left(\frac{(-1)^n}{n-1}+\frac{(-1)^n}{n+1}\right) \\
&= 2\ln 2 -1
\end{aligned}
$$
A: 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}
