State Jordan's Lemma.
Consider the integral
I=∮Cdz(a2+z2)sinπzzsin(xz)
for real x and a. The rectangular contour C runs from +∞+iϵ to −∞+iϵ, to −∞−iϵ, to +∞−iϵ and back to +∞+iϵ, where ϵ is infinitesimal and positive. Perform the integral in two ways to show that
n=−∞∑∞(−1)na2+n2nsinnx=−πsinhaπsinhax
for ∣x∣<π.