Solving real integrals by residue theorem

I have a concrete problem. I've solved a lot of real integrals, but I just can't solve this one:

$\displaystyle \int_0^\infty \frac{1}{(4x^2+\pi^2)*cosh(x)} dx $

I used rectangle path of integration (0, 0) -> (R,0), (R,0)->(R,2Pi*I), (R,2Pi*I)->(0,2Pi*I), R-->infinity. Residue (in 1/2*I*Pi) of this function by this path, or path of rectangle from -oo to +oo (I can do that since it is an even function, thus integral is 1/2 of value by this path), with height of 2Pi*I (or just Pi*I) is 0.

Integral is not a zero, infact the solution is ln2/(2Pi), so it means that probably integral from 2Pi*I to 0 on Im axis is not zero! I can't calculate that integral (and it seems to me that its zero!)... Anyway, If anyone knows how to solve this integral (and belive me, its not easy), please notify me as soon as you can, here, or by email: edeumic@xnet.hr. (You can solve it on a paper, scan it, and than send it if you find that easyer). Oh yeah, don't solve it numerically...

Thank you.