# Asymptotic expansion of Theta type sum

Hello

I am trying to find the asymptotic expansion for $t_1,t_2 \to 0+$ of the function

$$F_{w_0,\tau}(t_1,t_2) = \sum_{w \in \mathbb{Z}\tau+\mathbb{Z}} w \ \operatorname{exp}(- |w|^2t_1 - |w_0-w|^2t_2)$$

where we may assume that $w_0 \in \mathbb{Z}\tau +\mathbb{Z}$. The obvious first step would probably be to try and use the transformation properties of the theta functions, but this seems to lead to a rather complicated expression involving rational functions in the $t_i$. Hope anyone has suggestions/references/etc.

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You can use the Poisson summation formula. Your summand is essentially a Gaussian, $\sim \exp(-Q(m,n))$, where $w=m+\tau n$ and $Q$ is a quadratic polynomial in $m$ and $n$. The Poisson summation formula converts this sum into a sum with a different essentially Gaussian summand, $\sim \exp(-P(2\pi m,2\pi n))$, where $P$ is another quadratic polynomial (the Legendre transform of $Q$, which is what you get after when treating $m$ and $n$ as continuous variables and taking a Fourier transform).
Since, for small $t_1$ and $t_2$, you suppose $Q \sim O(t_1,t_2) O(m^2,n^2,mn)$, you'll also have $P \sim O(t_1^{-1},t_2^{-2}) O(m^2,n^2,mn)$. Therefore, the magnitude of the terms in the new sum will decrease as the tail of a Gaussian for successively higher values of $|m|$ and $|n|$. Essentially, the new series is itself an asymptotic expansion.
Is not the answer valid for the summand $\exp(...)$ rather than $w\exp(...)$ as in the question? –  მამუკა ჯიბლაძე Feb 9 at 9:47
Poisson summation requires the Fourier transform of the summand (with respect to $w$) in this case. Integration by parts easily changes $w \exp(\cdots)$ into $\exp(\cdots)$ under the Fourier integral. –  Igor Khavkine Feb 9 at 12:28