That sum is obtained from taking the residues inside a square contour with right boundary on [tex]c\pm i\infty[/tex] and allowed to expand to infinity. The integral tends to zero on the remaining three legs and we're left with (via the Residue Theorem) an infinite sum of residues corresponding to the poles at [tex]-n[/tex]:
Ok, approach this strictly in terms of contour integration: it's a rectangular contour with corners at [tex]c\pm i \infty[/tex], and [tex]-\infty\pm i\infty[/tex]. Then that contour encloses the poles at [tex]n=0,-1,-2,-3,\cdots[/tex]. And by the Residue Theorem, the integral, assuming the other legs go to zero, is just [tex]2\pi i[/tex] times the sum of the residues at that set of poles.
Also, I though it was a bit of a challenge to show: