Solved – Derivation of variance of normal distribution with gamma function

So I'm reading about the derivation of the variance for normal distribution and I don't understand the following derivation with the use of gamma function.

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So, if I continue this derivation the integral becomes

2int_{-infty}^infty ue^{-u}du

which is clearly not gamma function (in gamma function integral goes from 0 to infinity). How can I solve this integral?

Note that as $z^2=(-z)^2$ then $$dfrac{sigma^2}{sqrt{2pi}}int_{-infty}^infty z^2e^{-z^2/2}dz=2dfrac{sigma^2}{sqrt{2pi}}int_{0}^infty z^2e^{-z^2/2}dz,,$$ that is, as the mean is $z=0$, the function is symmetric around the $z=0$ axis, and the twice the $[0,infty)$ area is the $(-infty,infty)$ area.

Now let $u=z^2/2$, then $du=zdz$ and $dz=dfrac{du}{sqrt{2u}}$ and

$$2dfrac{sigma^2}{sqrt{2pi}}int_{0}^infty z^2e^{-z^2/2}dzrightarrow2dfrac{sigma^2}{sqrt{pi}}int_{0}^infty u^{1/2}e^{-u}du,.$$

Finally, $Gamma(z) = int_0^infty x^{z-1} e^{-x},dx$, thus our integral can be rewritten as $$2dfrac{sigma^2}{sqrt{pi}}Gammaleft(dfrac{3}{2}right)=sigma ^2,,.$$

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