True or false, normally?

Question

I understand that under the normal curve the area is 1. But numerically $\sum_{-\infty}^\infty e^{-k^2}=\sqrt{2\pi}$ holds with a very high precision. Do we have equality? If not, why are they so close? Note that $\sum_{-\infty}^\infty e^{-r k^2}=\sqrt{2\pi}$ is not true for $r=0.99,r=1.01$.

Answer 1

I assume that in your sum (which I think should be an integral), $k=\frac{x}{\sqrt{ 2 \sigma}} = \frac{x}{\sqrt 2}$, otherwise your sum comes out to $\sqrt \pi$, not $\sqrt{ 2\pi}$. Let's call your integral $S$.

Remember that $$f(x|\mu ,\sigma ^{2})={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}$$

With $\mu = 0$ and $\sigma^2=1$ (a typical normal distribution), we get that $$\int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^\infty \frac 1{\sqrt {2\pi}} e^{-k^2}=\frac 1{\sqrt {2\pi}} \int_{-\infty}^\infty e^{-k^2} = \frac 1{\sqrt {2\pi}}S$$ is $\frac 1{\sqrt {2\pi}}$ times what you found. And, of course, $$\frac 1{\sqrt {2\pi}} \cdot \sqrt{2\pi} = 1$$ as expected.

Answer 2

In terms of elliptic theta functions, we have $$ \vartheta \left(0;\frac{i}{\pi} \right)=\theta_3(0;\mathrm{e}^{-1})=\sum_{k=-\infty}^\infty \mathrm{e}^{-k^2}$$ and numerical evaluation gives $$\theta_3^2(0; \mathrm{e}^{-1}) \approx 3.1422426599356463391$$ this is not close to $2\pi$, yet a little close to $\pi$.