James Dale

Answered

2022-01-05

Prove:

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

Answer & Explanation

Bernard Lacey

Expert

2022-01-06Added 30 answers

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

$\frac{{d}^{n}}{d{a}^{n}}I\left(a\right)={(-1)}^{n}{\int}_{0}^{\mathrm{\infty}}{x}^{2n}{e}^{-a{x}^{2}}dx$

some algebraic manipulation and evaluating at$a=1$ will yield the wanted identity.

differentiate n times with respect to

To spell things a little more out, this technique is known as Differentiation under the integral sign. Using the fact that

some algebraic manipulation and evaluating at

godsrvnt0706

Expert

2022-01-07Added 31 answers

Alternatively, integration by parts works immediately.

$a}_{n}={\int}_{0}^{\mathrm{\infty}}{x}^{2n}{e}^{-{x}^{2}$

Consider$U={x}^{2n-1}$ so that $du=(2n-1){x}^{2n-2}$ , and $dv=x{e}^{-{x}^{2}}$ so that $V=-\frac{1}{2}{e}^{-{x}^{2}}$ .

Then

${\int}_{0}^{\mathrm{\infty}}{x}^{2n}{e}^{-{x}^{2}}dx=\frac{1}{2}{e}^{-{x}^{2}}{x}^{2n-1}{\mid}_{0}^{\mathrm{\infty}}-{\int}_{0}^{\mathrm{\infty}}(2n-1){x}^{2n-2}\frac{-1}{2}{e}^{-{x}^{2}}dx$

$=\frac{2n-1}{2}{\int}_{0}^{\mathrm{\infty}}{x}^{2n-2}{e}^{-{x}^{2}}dx=\frac{(2n-1)2n}{{2}^{2}n}{\int}_{0}^{\mathrm{\infty}}{x}^{2n-2}{e}^{-{x}^{2}}dx$

Hence

$a}_{n}=\frac{\left(2n\right)(2n-1)}{{2}^{2}n}{a}_{n-1$

and since$a}_{0}=\frac{\sqrt{\pi}}{2$ we conclude

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

Consider

Then

Hence

and since

star233

Expert

2022-01-11Added 238 answers

Let's suppose that, one way or another, you know that

On the other hand,

Finally, note that by evenness,

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