# A discrete version of the normal distribution

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
$$\displaystyle \int_{-\infty }^{\infty }\!1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2} }{\sigma\, \sqrt{\pi }}{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}\,{\rm d}x=\sum _{x=-\infty }^{\infty }1/2\,{\frac { \left( x-\mu \right) ^{2} \sqrt{2}}{\sigma\, \sqrt{\pi } }{{\rm e}^{-1/2\,{\frac { \left( x-\mu \right) ^{2}}{{\sigma}^{2}}}}}}$$

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