Gaussian-like integrals
Earlier I presented a classic proof of the Gaussian integral,
\begin{align}
\int\limits_{-\infty}^{\infty} \! e^{-x^2}\,\mathrm d x = \sqrt{\pi}\,.
\end{align}
In this post I show a formula concerning a class of similar integrals, namely
\begin{align}
\int\limits_{0}^{\infty} \! x^{n} e^{-x^2}\,\mathrm d x\,, \quad n\neq -1\,.
\end{align}
The formula in question is obtained using integration by parts. To recap, integration by parts allows recasting certiain integrals as follows:
\begin{align}
\int\limits_a^b f'(x)\cdot g(x)\,\mathrm dx = \Big[f(x)\cdot g(x)\Big]_a^b - \int\limits_a^b\! f(x) \cdot g'(x)\,\mathrm d x\,.
\end{align}
Letting $f'(x)=x^n$, so that $f(x)=\frac{1}{n+1} x^{n+1}$, and $g(x) = e^{-x^2}$ such that $g'(x)=-2x\cdot e^{-x^2}$ (using the chain rule), we obtain
\begin{align}
\int\limits_0^{\infty} \! x^n e^{-x^2}\,\mathrm d x &= \left[\frac{1}{n+1} x^{n+1}\cdot e^{-x^2}\right]_0^{\infty} - \int\limits_0^{\infty} \frac{1}{n+1} x^{n+1}\cdot \left(-2x\cdot e^{-x^2}\right)\,\mathrm d x\\
&= \frac{2}{n+1} \int\limits_0^{\infty}\! x^{n+2} e^{-x^2}\,\mathrm d x\,,
\end{align}
where the bracket-term vanishes in both limits.
Rearrranging gives the desired formula:
\begin{align}
\int\limits_0^{\infty} \! x^{n+2} e^{-x^2}\,\mathrm d x &= \frac{n+1}{2} \int\limits_0^{\infty} \! x^n e^{-x^2}\,\mathrm d x\,.
\end{align}
We will now use this for obtaining a number of integrals of the type mentioned above.
Using the Gaussian integral (and its symmetry around zero) we have that for the case of $n=0$, \begin{align} \int\limits_0^{\infty} \! x^2 e^{-x^2}\,\mathrm d x &= \frac{1}{2}\cdot \int\limits_0^{\infty}\! e^{-x^2}\,\mathrm d x = \frac{\sqrt{\pi}}{4}\,. \end{align} This is a useful result e.g. in kinetic gas theory, as shown here.
The formula allows us to obtain iteratively any value of the integral for even values of $n$. In the case of odd values of $n$ we consider the integral \begin{align} \int\limits_0^{\infty}\! x e^{-x^2} \,\mathrm d x\,, \end{align} which is easily obtained with the substitution $u = x^2$, \begin{align} \int\limits_0^{\infty}\! x e^{-x^2} \,\mathrm d x &= \frac{1}{2} \int\limits_0^{\infty}\! e^{-u}\,\mathrm d u\\ &= -\frac{1}{2}\left[e^{-u}\right]_0^{\infty}\\ &= \frac{1}{2}\,. \end{align} This is the input needed in our formula with $n=1$, yielding \begin{align} \int\limits_0^{\infty} x^{3} e^{-x^2}\,\mathrm d x &= \frac{1+1}{2}\int\limits_0^{\infty}\! x e^{-x^2}\,\mathrm d x \\ &= \frac{1}{2}\,, \end{align} and again for $n=3$, \begin{align} \int\limits_0^{\infty} x^{5} e^{-x^2}\,\mathrm d x &= \frac{1+3}{2}\int\limits_0^{\infty}\! x^3 e^{-x^2}\,\mathrm d x \\ &= \frac{5}{4}\,, \end{align} and so on.
Using the Gaussian integral (and its symmetry around zero) we have that for the case of $n=0$, \begin{align} \int\limits_0^{\infty} \! x^2 e^{-x^2}\,\mathrm d x &= \frac{1}{2}\cdot \int\limits_0^{\infty}\! e^{-x^2}\,\mathrm d x = \frac{\sqrt{\pi}}{4}\,. \end{align} This is a useful result e.g. in kinetic gas theory, as shown here.
The formula allows us to obtain iteratively any value of the integral for even values of $n$. In the case of odd values of $n$ we consider the integral \begin{align} \int\limits_0^{\infty}\! x e^{-x^2} \,\mathrm d x\,, \end{align} which is easily obtained with the substitution $u = x^2$, \begin{align} \int\limits_0^{\infty}\! x e^{-x^2} \,\mathrm d x &= \frac{1}{2} \int\limits_0^{\infty}\! e^{-u}\,\mathrm d u\\ &= -\frac{1}{2}\left[e^{-u}\right]_0^{\infty}\\ &= \frac{1}{2}\,. \end{align} This is the input needed in our formula with $n=1$, yielding \begin{align} \int\limits_0^{\infty} x^{3} e^{-x^2}\,\mathrm d x &= \frac{1+1}{2}\int\limits_0^{\infty}\! x e^{-x^2}\,\mathrm d x \\ &= \frac{1}{2}\,, \end{align} and again for $n=3$, \begin{align} \int\limits_0^{\infty} x^{5} e^{-x^2}\,\mathrm d x &= \frac{1+3}{2}\int\limits_0^{\infty}\! x^3 e^{-x^2}\,\mathrm d x \\ &= \frac{5}{4}\,, \end{align} and so on.
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