- The Poisson summation formula applied to specific examples often provides interesting identities.
- Let \(\tau\) be fixed with \(\operatorname { I m } ( \tau ) > 0\) . Apply the Poisson summation formula to
\[ f (z) = (\tau + z) ^ {- k}, \]
where k is an integer \(\geq 2 .\) , to obtain
\[ \sum_ {n = - \infty} ^ {\infty} \frac {1}{(\tau + n) ^ {k}} = \frac {(- 2 \pi i) ^ {k}}{(k - 1) !} \sum_ {m = 1} ^ {\infty} m ^ {k - 1} e ^ {2 \pi i m \tau}. \]
- Set \(k = 2\) in the above formula to show that if \(\operatorname { I m } ( \tau ) > 0\) , then
\[ \sum_ {n = - \infty} ^ {\infty} \frac {1}{(\tau + n) ^ {2}} = \frac {\pi^ {2}}{\sin^ {2} (\pi \tau)}. \]
- Can one conclude that the above formula holds true whenever \(\tau\) is any complex number that is not an integer?
[Hint: For (a), use residues to prove that \(\hat { f } ( \xi ) = 0 , \mathrm { i f } \ \xi < 0\) , and
\[ \hat {f} (\xi) = \frac {(- 2 \pi i) ^ {k}}{(k - 1) !} \xi^ {k - 1} e ^ {2 \pi i \xi \tau}, \quad \text { when } \xi > 0. ] \]
Solution
Let \[ f(x)=(\tau+x)^{-k},\qquad \Im\tau>0,\quad k\ge2. \] The only pole of \(f(z)\) is at \(z=-\tau\), which lies in the lower half-plane.
For \(\xi<0\), close the contour in the upper half-plane. The exponential \(e^{-2\pi iz\xi}\) decays there and there are no poles, so \[ \widehat f(\xi)=0. \] For \(\xi>0\), close in the lower half-plane. The contour is clockwise, and the pole at \(-\tau\) has order \(k\). Its residue is \[ \frac1{(k-1)!} \left.\frac{d^{k-1}}{dz^{k-1}}e^{-2\pi iz\xi}\right|_{z=-\tau} =\frac{(-2\pi i\xi)^{k-1}}{(k-1)!}e^{2\pi i\tau\xi}. \] Thus \[ \widehat f(\xi) =-2\pi i\,\operatorname{Res}_{z=-\tau} \frac{e^{-2\pi iz\xi}}{(z+\tau)^k} =\frac{(-2\pi i)^k}{(k-1)!}\xi^{k-1}e^{2\pi i\tau\xi}. \] At \(\xi=0\), closing in the upper half-plane gives \(\widehat f(0)=0\).
Poisson summation now yields \[ \sum_{n\in\mathbb Z}\frac1{(\tau+n)^k} =\sum_{m\in\mathbb Z}\widehat f(m) =\frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty}m^{k-1}e^{2\pi im\tau}. \] This proves (a).
For \(k=2\), let \(q=e^{2\pi i\tau}\); since \(\Im\tau>0\), \(|q|<1\). Then \[ \sum_{m=1}^{\infty}m q^m=\frac{q}{(1-q)^2}, \] so \[ \sum_{n\in\mathbb Z}\frac1{(\tau+n)^2} =-4\pi^2\frac{q}{(1-q)^2}. \] But \[ \sin^2(\pi\tau)=-\frac{(1-q)^2}{4q}, \] hence \[ \sum_{n\in\mathbb Z}\frac1{(\tau+n)^2} =\frac{\pi^2}{\sin^2(\pi\tau)}. \]
The identity remains valid for every \(\tau\in\mathbb C\setminus\mathbb Z\). Indeed, the series on the left converges normally on compact subsets of \(\mathbb C\setminus\mathbb Z\), so it defines a holomorphic function there; the right side is also holomorphic there. They agree on the upper half-plane, and \(\mathbb C\setminus\mathbb Z\) is connected, so the identity theorem extends the equality to all nonintegral \(\tau\).