Prove: \int_0^{\infty} x^{2n}e^{-x^2}dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}

James Dale 2022-01-05 Answered
Prove:
\(\displaystyle{\int_{{0}}^{{\infty}}}{x}^{{{2}{n}}}{e}^{{-{x}^{{2}}}}{\left.{d}{x}\right.}={\frac{{{\left({2}{n}\right)}!}}{{{2}^{{{2}{n}}}{n}!}}}{\frac{{\sqrt{{\pi}}}}{{{2}}}}\)

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Expert Answer

Bernard Lacey
Answered 2022-01-06 Author has 97 answers
Alternatively, set
\(\displaystyle{I}{\left(\alpha\right)}={\int_{{0}}^{\infty}}{e}^{{-{a}{x}^{{2}}}}{\left.{d}{x}\right.}\)
differentiate n times with respect to \(\displaystyle\alpha\) and evaluate at \(\displaystyle\alpha={1}\)
To spell things a little more out, this technique is known as Differentiation under the integral sign. Using the fact that \(\displaystyle{I}{\left(\alpha\right)}={\frac{{{1}}}{{{2}}}}\sqrt{{{\frac{{\pi}}{{\alpha}}}}}\) and differentiating to obtain
\(\displaystyle{\frac{{{d}^{{n}}}}{{{d}{a}^{{n}}}}}{I}{\left({a}\right)}={\left(-{1}\right)}^{{n}}{\int_{{0}}^{\infty}}{x}^{{{2}{n}}}{e}^{{-{a}{x}^{{2}}}}{\left.{d}{x}\right.}\)
some algebraic manipulation and evaluating at \(\displaystyle{a}={1}\) will yield the wanted identity.
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godsrvnt0706
Answered 2022-01-07 Author has 420 answers
Alternatively, integration by parts works immediately.
\(\displaystyle{a}_{{n}}={\int_{{0}}^{\infty}}{x}^{{{2}{n}}}{e}^{{-{x}^{{2}}}}\)
Consider \(\displaystyle{U}={x}^{{{2}{n}-{1}}}\) so that \(\displaystyle{d}{u}={\left({2}{n}-{1}\right)}{x}^{{{2}{n}-{2}}}\), and \(\displaystyle{d}{v}={x}{e}^{{-{x}^{{2}}}}\) so that \(\displaystyle{V}=-{\frac{{{1}}}{{{2}}}}{e}^{{-{x}^{{2}}}}\).
Then
\(\displaystyle{\int_{{0}}^{\infty}}{x}^{{{2}{n}}}{e}^{{-{x}^{{2}}}}{\left.{d}{x}\right.}={\frac{{{1}}}{{{2}}}}{e}^{{-{x}^{{2}}}}{x}^{{{2}{n}-{1}}}{{\mid}_{{0}}^{\infty}}-{\int_{{0}}^{\infty}}{\left({2}{n}-{1}\right)}{x}^{{{2}{n}-{2}}}{\frac{{-{1}}}{{{2}}}}{e}^{{-{x}^{{2}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{2}{n}-{1}}}{{{2}}}}{\int_{{0}}^{\infty}}{x}^{{{2}{n}-{2}}}{e}^{{-{x}^{{2}}}}{\left.{d}{x}\right.}={\frac{{{\left({2}{n}-{1}\right)}{2}{n}}}{{{2}^{{2}}{n}}}}{\int_{{0}}^{{\infty}}}{x}^{{{2}{n}-{2}}}{e}^{{-{x}^{{2}}}}{\left.{d}{x}\right.}\)
Hence
\(\displaystyle{a}_{{n}}={\frac{{{\left({2}{n}\right)}{\left({2}{n}-{1}\right)}}}{{{2}^{{2}}{n}}}}{a}_{{{n}-{1}}}\)
and since \(\displaystyle{a}_{{0}}={\frac{{\sqrt{{\pi}}}}{{{2}}}}\) we conclude
\(\displaystyle{a}_{{n}}={\int_{{0}}^{\infty}}{x}^{{{2}{n}}}{e}^{{-{x}^{{2}}}}{\left.{d}{x}\right.}={\frac{{{\left({2}{n}\right)}!}}{{{2}^{{{2}{n}}}{n}!}}}{\frac{{\sqrt{{\pi}}}}{{{2}}}}\)
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star233
Answered 2022-01-11 Author has 0 answers

Let's suppose that, one way or another, you know that \(\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}\). Then
\(\int_{-\infty}^\infty e^{2tx-x^2}dx=e^{t^2}\int_{-\infty}^\infty e^{-(t-x)^2}dx=e^{t^2}\sqrt{\pi}\)
On the other hand,
\(\int_{-\infty}^\infty e^{2tx-x^2}dx=\int_{-\infty}^{\infty}(\sum_{n\geq0}\frac{2^nt^nx^n}{n!})e^{-x^2}dx=\sum_{n\geq0}\frac{2^nt^n}{n!}\int_{-\infty}^{\infty}x^ne^{-x^2}dx\)
Finally, note that by evenness,
\(\int_{-\infty}^\infty x^{2n}e^{-x^2}dx=2\int_0^\infty x^{2n}e^{-x^2}dx\)

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