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Edit:

It is well known that for real variable $x$, $\lim_{x \to 1^-} (1 - x) \Big( \sum_{m = 0}^\infty x^{m^2} \Big)^2 = \frac{\pi}{4}$.

I would like to know whether this limit continues to hold if we instead let our variable $z$ be in $\mathbb{C}$. That is, do we have

$$\lim_{|z| < 1, \, z \to 1} (1 - z) \Big( \sum_{m = 0}^\infty z^{m^2} \Big)^2 = \frac{\pi}{4}?$$

The most accessible proof I know of for the real case essentially utilizes the integral test for series. So we cannot directly apply it the complex case because the integral test requires a nonnegative, decreasing function as input.

The main problem seems to be oscillation in the sense that, even as $z \to 1$, the powers $z^{m^2}$ can have phases that are quite different from one another.

However, note that in the simpler case of a geometric series $\sum_{m =0}^\infty z^m = (1 -z)^{-1}$ these oscillations do not cause an issue, in the sense that $\lim_{|z|<1,\, z \to 1} (1 -z) \sum_{m =0}^\infty z^m = 1$. So I thought maybe there is some hope.

We can try to bootstrap back the real case:

\begin{equation*} \begin{split} \Big| \sqrt{1 - z}\sum_{m = 0}^\infty z^{m^2} -\frac{\sqrt{\pi}}{2} \Big| &\le \Big| \sqrt{1 - x}\sum_{m = 0}^\infty x^{m^2} -\frac{\sqrt{\pi}}{2} \Big| \\ &+ \Big| \sqrt{1 - x}\sum_{m = 0}^\infty x^{m^2} - \sqrt{1 - z}\sum_{m = 0}^\infty z^{m^2} \Big|, \end{split} \end{equation*} but it seems quite tricky to show that the last expression is small uniformly in the phase of $z$.

Original post:

I would like to understand the asymptotic for the series $$\sum_{m =1}^\infty e^{-m^2/(\sigma + it)},$$ for $0 < \sigma \le 1$ and $ t \to \infty$. This series is similar to but a little different from the series representation for Jacobi's third theta function. The most helpful reference I have been able to find so far is this previous post, but the asymptotic quoted there does not quite apply to my situation.

If one replaces $m^2$ by $m$ then the asymptotic can be found immediately because in that case we just have a geometric series

$$\sum_{m =0}^\infty e^{-m/(\sigma + it)} = \frac{e^{-1/(\sigma + it)}}{1 - e^{-1/(\sigma + it)}} \sim (\sigma + it), \qquad 0 < \sigma \le 1, \, t \to \infty.$$

My guess is that having $m^2$ instead of $m$ will give $$\sum_{m =1}^\infty e^{-m^2/(\sigma + it)} = O(|\sigma + it|^{1/2}).$$

I am familiar with the Poisson summation formula and Jacobi's theta function identity. I'm not sure if they are helpful here. Hints, references, or solutions are greatly appreciated.

JZS
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    your function is given by $2f(s)+1=\theta_0(-\frac{1}{i\pi s})$ with $s=\sigma+it, \theta_0(z)=\sum_{n \in \mathbb Z}e^{in^2\pi z}, \Im z >0$ so applying the usual inversion formula $\theta_0(z)=\frac{e^{i\pi/4}}{\sqrt z}\theta_0(-1/z)$ should give you the result you want easily – Conrad Nov 27 '24 at 17:22
  • @Conrad Thank you for your comment. I had considered taking a route similar to this, but I think the catch is that it's not immediately clear what happens to $\theta_0(-1/z)$ as $\sigma \to 0^+$ and $t \to \infty$. – JZS Nov 27 '24 at 17:25
  • @Conrad Thanks again for the comment. I think I'm getting closer, but then don't we have $f(s) = \tfrac{1}{2} \theta_0(-\tfrac{1}{i\pi s}) - \tfrac{1}{2}$, which is $\tfrac{e^{i\pi/4} \sqrt{-i \pi s}}{2} \theta_0(i\pi s) - \tfrac{1}{2}$. So I'm getting that the relevant sum is effectively $\sum e^{n^2(-\sigma + it)}$, which is troublesome as $\sigma \to 0^+$. – JZS Nov 27 '24 at 18:00
  • @Conrad That multiplication makes sense, but then substituting $z = i \pi s = \pi(-t + i\sigma)$ into $\theta_0(z) = \sum_{n \in \mathbb{Z}} e^{in^2 \pi z}$ is where I find trouble, because there is yet another $i$ factor, before the $n^2$ in the exponent, by which I need to multiply. – JZS Nov 27 '24 at 18:11
  • yes you are right and I was hasty - I guess that then you need to fix $\sigma$ or at least get the asymptotic for $\sigma \ge \delta>0$ and the $O$ depending on $\delta$ – Conrad Nov 27 '24 at 18:29

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