Evaluation of $\sum_{x=0}^\infty e^{-x^2}$

Question

Most of us are aware of the classic Gaussian Integral

$$\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$

I would be interested in evaluating the similar sum

$$\sum_{x=0}^\infty e^{-x^2}$$

Now, because $\exp(-\lfloor x \rfloor^2) \ge \exp(-x)$, we find

$$\sum_{x=0}^\infty e^{-x^2}= \int_0^\infty e^{-\lfloor x \rfloor^2}\, dx \ge \int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$

Does a closed form for this sum exist? If so, what would it be? I would be very interested in how a closed form would be found for this function.

Answer 1

As mentioned in the comments, this sum is related to one of the Jacobi theta functions. $$\begin{eqnarray*} \sum_{n=0}^\infty e^{-n^2} &=& \frac{1}{2}\left(1 + \sum_{n=-\infty}^\infty \left(\frac{1}{e}\right)^{n^2}\right) \\ &=& \frac{1}{2}\left[1+\vartheta_3\left(0,\frac{1}{e}\right)\right] \\ &\simeq& 1.386 \end{eqnarray*}$$