This is a challenge problem I already know how to solve.

Show via the Residue Theorem and without using Bessel functions or any of that other stuff in the other thread:

$\displaystyle \int_{-1}^{1}\frac{\cosh(2z)}{\sqrt{1-z^2}}dz=\pm\pi \sum_{k=0}^{\infty} \frac{1}{(k!)^2}$

with the $\displaystyle \pm$ due to the particular sheet of the underlying Riemann surface the integral is being integrated over.