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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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    You might want to investigate "theta functions". You're taking a standard one and evaluating it at $e$, and I doubt any good will come of that.2012-07-25
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    Curiously, though, replacing $e^{-n^2}$ with $e^{-\pi n^2}$ _does_ yield an explicit value. The best analogy might be to trigonometric functions; in effect asking for $\sum e^{-n^2}$ is similar to asking for $\sin(1)$.2012-07-25
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    @GerryMyerson Is it Bell-shaped summation?2012-07-25
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    @Frank, I don't know what is meant by "Bell-shaped summation".2012-07-25
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    @Steven, yes, the replacement yields an explicit value - which involves $\Gamma(1/4)$. If one accepts that as a closed form, maybe one also accepts answers involving theta functions?2012-07-25
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    @GerryMyerson Graham, Knuth, Patashnik: *Concrete Mathematics*, section 9.6, summation 4: $\Theta_n=\sum_k\exp(-k^2/n)$. Unfortunately, that book only gives out an asymptotic value: $\Theta_n/\sqrt{\pi n}=1+2\exp(-n\pi^2)+O(\exp(-4n\pi^2))$. I don't know why it's called *Bell-shaped summand*, but it's related to the Bell number $\varpi_n=e^{-1}\sum_{k\ge0}k^n/k!$.2012-07-25
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    @Frank, no, it's got nothing to do with Bell numbers. I does have something to do with the *bell curve* (capitalization matters!)...2012-07-25
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    @J.M. aha, it's just *bell-shaped* **summand**, not **summation**.2012-07-26

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