James Dale

2022-01-05

Prove:
${\int }_{0}^{\mathrm{\infty }}{x}^{2n}{e}^{-{x}^{2}}dx=\frac{\left(2n\right)!}{{2}^{2n}n!}\frac{\sqrt{\pi }}{2}$

Bernard Lacey

Expert

Alternatively, set
$I\left(\alpha \right)={\int }_{0}^{\mathrm{\infty }}{e}^{-a{x}^{2}}dx$
differentiate n times with respect to $\alpha$ and evaluate at $\alpha =1$
To spell things a little more out, this technique is known as Differentiation under the integral sign. Using the fact that $I\left(\alpha \right)=\frac{1}{2}\sqrt{\frac{\pi }{\alpha }}$ and differentiating to obtain
$\frac{{d}^{n}}{d{a}^{n}}I\left(a\right)={\left(-1\right)}^{n}{\int }_{0}^{\mathrm{\infty }}{x}^{2n}{e}^{-a{x}^{2}}dx$
some algebraic manipulation and evaluating at $a=1$ will yield the wanted identity.

godsrvnt0706

Expert

Alternatively, integration by parts works immediately.
${a}_{n}={\int }_{0}^{\mathrm{\infty }}{x}^{2n}{e}^{-{x}^{2}}$
Consider $U={x}^{2n-1}$ so that $du=\left(2n-1\right){x}^{2n-2}$, and $dv=x{e}^{-{x}^{2}}$ so that $V=-\frac{1}{2}{e}^{-{x}^{2}}$.
Then
${\int }_{0}^{\mathrm{\infty }}{x}^{2n}{e}^{-{x}^{2}}dx=\frac{1}{2}{e}^{-{x}^{2}}{x}^{2n-1}{\mid }_{0}^{\mathrm{\infty }}-{\int }_{0}^{\mathrm{\infty }}\left(2n-1\right){x}^{2n-2}\frac{-1}{2}{e}^{-{x}^{2}}dx$
$=\frac{2n-1}{2}{\int }_{0}^{\mathrm{\infty }}{x}^{2n-2}{e}^{-{x}^{2}}dx=\frac{\left(2n-1\right)2n}{{2}^{2}n}{\int }_{0}^{\mathrm{\infty }}{x}^{2n-2}{e}^{-{x}^{2}}dx$
Hence
${a}_{n}=\frac{\left(2n\right)\left(2n-1\right)}{{2}^{2}n}{a}_{n-1}$
and since ${a}_{0}=\frac{\sqrt{\pi }}{2}$ we conclude
${a}_{n}={\int }_{0}^{\mathrm{\infty }}{x}^{2n}{e}^{-{x}^{2}}dx=\frac{\left(2n\right)!}{{2}^{2n}n!}\frac{\sqrt{\pi }}{2}$

star233

Expert

Let's suppose that, one way or another, you know that ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{x}^{2}}dx=\sqrt{\pi }$. Then
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2tx-{x}^{2}}dx={e}^{{t}^{2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\left(t-x{\right)}^{2}}dx={e}^{{t}^{2}}\sqrt{\pi }$
On the other hand,
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2tx-{x}^{2}}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(\sum _{n\ge 0}\frac{{2}^{n}{t}^{n}{x}^{n}}{n!}\right){e}^{-{x}^{2}}dx=\sum _{n\ge 0}\frac{{2}^{n}{t}^{n}}{n!}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{n}{e}^{-{x}^{2}}dx$
Finally, note that by evenness,
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2n}{e}^{-{x}^{2}}dx=2{\int }_{0}^{\mathrm{\infty }}{x}^{2n}{e}^{-{x}^{2}}dx$

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