Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 24 / 4 / i / Solution

Use the Fourier transform convention $\hat{h}(\xi )={\int }_{\mathbb{R}}h(x)e^{-2\pi ix\xi }dx$. For a Schwartz function $h$, the Poisson summation formula isTo prove it, periodize $h$ as $P(x)={\sum }_{n\in \mathbb{Z}}h(x+n)$. Rapid decay of all derivatives makes this a smooth periodic function, with uniform termwise differentiability. Its $m$th Fourier coefficient iswhere absolute convergence justifies splitting the real-line integral into unit intervals. Integration by parts gives rapid decay of the coefficients of $P$, so its Fourier series converges absolutely and uniformly to $P$. Evaluate it at zero to obtain the formula.

For $\mathrm{Im}\tau >0$ and $z\in \mathbb{C}$, take the complex Gaussian function $h(x)=e^{\pi i\tau x^2+2\pi izx}$. It is a Schwartz function of the real variable $x$. Its Fourier transform isThe real Gaussian integral proves this first for $\tau =iy$, $y>0$, and real $z$. Normal convergence of the defining integrals makes both sides entire in $z$ and holomorphic in $\tau \in \mathbb{H}$, so the identity theorem extends it to the stated domain. Choose the square-root branch holomorphic for $-i\tau$ in the right half-plane and positive when $\tau =iy$.

Applying Poisson summation gives the transformation of the Jacobi theta function:The series on the right is ${\vartheta }_{00}(z/\tau ,-1/\tau )$. Rearranging therefore givesThe branch choice is part of the formula; at $\tau =i$ its square-root factor equals one.