Here is a fun looking integral.
$\int_{0}^{\infty}\frac{1}{(4x^{2}+{\pi}^{2})\cosh(x)}dx=\frac{\ln(2)}{2\pi}$.
I rewrote it as $\frac{2e^{z}}{(4z^{2}+{pi}^{2})(e^{2z}+1)}$
It would appear there is a pole of order 2 at $\frac{\pi i}{2}$.
This is due to it being a zero of cosh and the rational part.
I think the residue at $\frac{\pi i}{2}$ is $\frac{-i}{4{\pi}^{2}}$. I found this by looking at the Laurent expansion.
Thus, $2\pi i(\frac{-i}{4\pi^{2}})=\frac{1}{2\pi}$
I thought about using a rectangular contour with vertices $-R,R,R+\pi i, -R+\pi i$that
encloses $\frac{\pi i}{2}$ because there are an infinite number of poles due to the poles
of cosh being $\frac{(2n+1)\pi i}{2}$.
So, the only pole enclosed by C would be $\frac{\pi i}{2}$.
Now, how can I deal with this?.
For the top horizontal , I tried $\frac{2e^{x+\pi i}}{(4(x+\pi i)^{2}+{\pi}^{2})(e^{2(x+\pi i)}+1)}=\frac{-2e^{x}}{(4(x+\pi i)^{2}+{\pi}^{2})(e^{2x}+1)}$
But, this looks kind of messy with that square in the denominator.
Does anyone have any ideas on how to arrive at that ln(2)?. Thanks much.