Question 7.25: Let f(z) be such that along the path CN of Fig. 7-13, |f(z)|...
Let f(z) be such that along the path C_{N} of Fig. 7-13, |f(z)| \leq M /|z|^{k} where k>1 and M are constants independent of N. Prove that
\sum\limits_{-\infty}^{\infty} f(n)=-\{\text { sum of residues of } \pi \cot \pi z f(z) \text { at the poles of } f(z)\}
Case 1: f(z) has a finite number of poles.
In this case, we can choose N so large that the path C_{N} of Fig. 7-13 encloses all poles of f(z). The poles of \cot \pi z are simple and occur at z=0, \pm 1, \pm 2, \ldots
Residue of \pi \cot \pi z f(z) at z=n, n=0, \pm 1, \pm 2, \ldots, is
\underset{z \rightarrow n}{\lim}(z-n) \pi \cot \pi z f(z)=\underset{z \rightarrow n}{\lim} \pi\left(\frac{z-n}{\sin \pi z}\right) \cos \pi z f(z)=f(n)
using L’Hospital’s rule. We have assumed here that f(z) has no poles at z=n, since otherwise the given series diverges.
By the residue theorem,
\oint\limits_{C_{N}} \pi \cot \pi z f(z) d z=\sum\limits_{n=-N}^{N} f(n)+S (1)
where S is the sum of the residues of \pi \cot \pi z f(z) at the poles of f(z). By Problem 7.24 and our assumption on f(z), we have
\left|\oint\limits_{C_{N}} \pi \cot \pi z f(z) d z\right| \leq \frac{\pi A M}{N^{k}}(8 N+4)
since the length of path C_{N} is 8N + 4. Then, taking the limit as N → ∞, we see that
\underset{N \rightarrow \infty}{\lim} \oint\limits_{C_{N}} \pi \cot \pi z f(z) d z=0 (2)
Thus, from (1) we have as required,
\sum\limits_{-\infty}^{\infty} f(n)=-S (3)
Case 2: f(z) has infinitely many poles.
If f(z) has an infinite number of poles, we can obtain the required result by an appropriate limiting procedure. See Problem 7.103.