Editing Gaussian integral (section)

===Integrals of similar form===
<math display="block">\int_0^\infty x^{2n}  e^{-{x^2}/{a^2}}\,dx = \sqrt{\pi}\frac{a^{2n+1} (2n-1)!!}{2^{n+1}}</math>
<math display="block">\int_0^\infty x^{2n+1} e^{-{x^2}/{a^2}} \, dx = \frac{n!}{2} a^{2n+2}</math>
<math display="block">\int_0^\infty x^{2n}e^{-bx^2}\,dx = \frac{(2n-1)!!}{b^n 2^{n+1}} \sqrt{\frac{\pi}{b}}</math>
<math display="block">\int_0^\infty x^{2n+1}e^{-bx^2}\,dx = \frac{n!}{2b^{n+1}}</math>
<math display="block">\int_0^\infty x^{n}e^{-bx^2}\,dx = \frac{\Gamma(\frac{n+1}{2})}{2b^{\frac{n+1}{2}}}</math>
where <math>n</math> is a positive integer

An easy way to derive these is by [[Leibniz integral rule#Evaluating definite integrals|differentiating under the integral sign]].

<math display="block">\begin{align}
\int_{-\infty}^\infty x^{2n} e^{-\alpha x^2}\,dx
&= \left(-1\right)^n\int_{-\infty}^\infty \frac{\partial^n}{\partial \alpha^n} e^{-\alpha x^2}\,dx \\[1ex]
&= \left(-1\right)^n\frac{\partial^n}{\partial \alpha^n} \int_{-\infty}^\infty e^{-\alpha x^2}\,dx\\[1ex]
&= \sqrt{\pi} \left(-1\right)^n\frac{\partial^n}{\partial \alpha^n}\alpha^{-\frac{1}{2}} \\[1ex]
&= \sqrt{\frac{\pi}{\alpha}}\frac{(2n-1)!!}{\left(2\alpha\right)^n}
\end{align}</math>

One could also integrate by parts and find a [[recurrence relation]] to solve this.