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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.

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    @J.M. aha, it's just *bell-shaped* **summand**, not **summation**.2012-07-26

1 Answers 1

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