TITLE: A tough integral $\int_0^{\infty}\frac{\operatorname{sech}(\pi x)}{1+4x^2}\, \mathrm dx $
QUESTION [13 upvotes]: I was working on an integral which I found on Quora. I simplified it a lot and ended up with this intgeral
$$\int_0^{\infty}\dfrac{\operatorname{sech}(\pi x)}{1+4x^2}\, \mathrm dx $$
I tried converting this into exponential form and using geometric series which ended up in this.
$$2\sum_{k=0}^\infty (-1)^k\int_0^\infty \dfrac{e^{-(2k+1)\pi x}}{4x^2+1}\, \mathrm dx $$
I didn't try to solve this using exponential integral, as I am not that much familiar with it.
Using Wolfram|Alpha, I figured out that this integral is equal to $\frac{1}{2}\ln{2}$. The simple answer makes me suspect if the integral is just a tricky one.
How can I evaluate this integral, using this method or any other method, except contour integration?

REPLY [6 votes]: A short solution by using the series of $\operatorname{sech}(x)$.
Since $$\operatorname{sech}(x)= 4\sum_{k=0}^{\infty}\frac{(-1)^k(2k+1)\pi}{(2k+1)^2\pi^2+4x^2}$$ and then we have $$\int_0^{\infty}\frac{\operatorname{sech}(\pi x)}{1+4x^2}dx=\frac{4}{\pi}\sum_{k=0}^{\infty}{(-1)^k(2k+1)}\int_0^{\infty}\frac{dx}{(m^2+4x^2)(1+4x^2)}=$$ where $m=2k+1$ and by partial fraction decomposition  we easily can deduce the integral $$\int_0^{\infty}\frac{dx}{(m^2+4x^2)(1+4x^2)}=\frac{\pi}{4m(m+1)}=\frac{\pi}{4(2k+1)(2k+2)}$$ and hence we have $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2(k+1)}=\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}=\frac{\log(2)}{2}$$