As [@TerryTao][1] stated correctly in the comments, a bound on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially: First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have $$ \theta(t) = \sqrt \frac \pi \gamma \sum_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta_3(it, e^{-\frac{\pi^2}{\gamma}}). $$ Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have $$ \frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta_3(z,q)}. $$ In the paper *A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165-183*, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$: $$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (-1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)} $$ From the value $|k|$ in the exponent in the definition of $a_k$ it follows directly that the $a_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a_k$ gives you the desired map $f$ so that $|a_k| \leq f(k)$. To derive the above formula, Jenssen refers to an exercise in the book *Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library)*. He essentially solves this exercise which leads to the above expression. The above formula should agree with the one derived in the answer of [@მამუკა ჯიბლაძე][2]. [1]: https://mathoverflow.net/users/766/terry-tao [2]: https://mathoverflow.net/users/41291/%E1%83%9B%E1%83%90%E1%83%9B%E1%83%A3%E1%83%99%E1%83%90-%E1%83%AF%E1%83%98%E1%83%91%E1%83%9A%E1%83%90%E1%83%AB%E1%83%94