A discrete version of the normal distribution

[Ad VanderVen]

TL;DR Summary

How can the result of an integral of a normal distribution be the same as the result of a sum?

I have the following function for the normal distribution:
$$\displaystyle f \left(x \right) = \frac{1}{2}~\frac{\sqrt{2}~e^{-\frac{1}{2}~\frac{\left(x -\mu \right)^{2}}{\sigma ^{2}}}}{\sigma ~\sqrt{\pi }}$$
How can the following integrals be equal to their sums?
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac { \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}},$$
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac {x \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac {x \sqrt{2}}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$
and
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2}
}{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2}}{\sigma\, \sqrt{\pi }
}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$

 

[mathman]

Looks like the integral and first two moments of a normal distribution.

 

[pbuk]

This appears to be almost a repost of https://www.physicsforums.com/threa...iscrete-random-variable.1013035/#post-6611188.

The fact that the integral and the infinite sum are equal is perhaps surprising, but it is a well-known result that a normal distribution is the limit of a binomial distribution as $n \to \infty$.

See for example https://openpress.usask.ca/introtoa...ibution-as-a-limit-of-binomial-distributions/