Integration of Maxwell-Boltzmann velocity distribution function

Question

How to show that the integration: $$\int_0^{\infty} x^ne^{-bx^2}\,dx=\left(\frac1b\right)^{(n+1)/2}\cdot {\Gamma\left(\frac{n+1}{2}\right) \over 2}$$

Answer 1

First we recall the definition of the Gamma function: $$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x}\,dx.$$

To simplify $$\int_0^{\infty} x^ne^{-bx^2}\,dx$$ we use the change of variables

\begin{align*} u &= bx^{2} \implies x = \sqrt{\frac{u}{b}}\\ du &= 2bx\,dx = 2b\sqrt{\frac{u}{b}}\,dx \implies dx = \frac{1}{2b}\left(\frac{u}{b}\right)^{-1/2}. \end{align*}

This gives us \begin{align*} \int_{0}^{\infty}x^{n}e^{-bx^{2}}\,dx &= \frac{1}{2b}\int_{0}^{\infty}\left(\frac{u}{b}\right)^{-1/2}\left(\frac{u}{b}\right)^{n/2}e^{-u}\,du\\[5pt] &=\frac{1}{2b}\int_0^\infty \left(\frac{u}{b}\right)^{\frac{n-1}{2}}e^{-u}\,du\\[5pt] &=\frac{1}{2}\left(\frac{1}{b}\right)^{(n+1)/2}\int_0^{\infty}u^{\frac{n+1}{2}-1}e^{-u}\,du\\[5pt] &= \left(\frac{1}{b}\right)^{(n+1)/2}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)}{2}. \end{align*}