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Use the following integral to find the first and second moments. The mean is, of course, the first moment and the variance is the second moment minus the first moment squared. :<math>\int_{0}^{\infty} x^{n} e^{-ax^2}\,\mathrm{d}x =
\begin{cases}
\frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\
\frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{integer}, a>0) \\
\frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{integer}, a>0)
\end{cases} </math> (!! is the [[Double factorial]])
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