# Please show $\int_0^\infty x^{2n} e^{-x^2}\mathrm dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$ without gamma function?

Prove:

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

Thanks!

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Please don't post questions as commands ("Prove ____." "Show _____.") because it is considered rude. –  Tyler Mar 14 '11 at 17:00

Alternatively, integration by parts works immediately.
Let $$a_n=\int_0^\infty x^{2n}e^{-x^2}.$$ Consider $U=x^{2n-1}$ so that $du=(2n-1)x^{2n-2}$, and $dv=xe^{-x^{2}}$ so that $V=-\frac{1}{2}e^{-x^{2}}$.

Then $$\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{1}{2}e^{-x^{2}}x^{2n-1}\biggr|_{0}^{\infty}-\int_{0}^{\infty}(2n-1)x^{2n-2}\frac{-1}{2}e^{-x^{2}}dx$$ $$=\frac{(2n-1)}{2}\int_{0}^{\infty}x^{2n-2}e^{-x^{2}}dx=\frac{(2n-1)2n}{2^{2}n}\int_{0}^{\infty}x^{2n-2}e^{-x^{2}}dx$$

Hence $$a_n=\frac{(2n)(2n-1)}{2^2n}a_{n-1}$$ and since $a_0=\frac{\sqrt{\pi}}{2}$ we conclude $$a_n=\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$$ by induction.

Hope that helps,

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Alternatively, set $$I(\alpha) = \int_0^\infty e^{-\alpha x^2}\mathrm{d}x,$$ differentiate $n$ times with respect to $\alpha$ and evaluate at $\alpha = 1$.

EDIT: To spell things a little more out, this technique is known as Differentiation under the integral sign. Using the fact that $I(\alpha) =\frac12\sqrt{\frac{\pi}{\alpha}}$ and differentiating to obtain $$\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n} I(\alpha) = (-1)^n\int_0^\infty x^{2n} e^{-\alpha x^2}\mathrm{d}x,$$ some algebraic manipulation and evaluating at $\alpha = 1$ will yield the wanted identity.

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Can you spell this out a little more? –  Grumpy Parsnip Mar 14 '11 at 16:55
Very nice answer! –  JavaMan Mar 14 '11 at 20:11
@Raeder I don't get it.... how to calculate $\frac{\text{d}^n}{\text{d}\alpha^n}I(\alpha)$? –  athos Aug 1 at 9:02

Making a change of variable $u=x^2$ gives $$\int_0^\infty {x^{2n} e^{ - x^2 } dx} = \frac{1}{2}\int_0^\infty {u^{n - 1/2} e^{ - u} du} = \frac{1}{2}\Gamma (n + 1/2).$$ Then from the well-known formula for the gamma function $$\Gamma (n + 1/2) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi$$ we get $$\int_0^\infty {x^{2n} e^{ - x^2 } dx} = \frac{{(2n)!}}{{2^{2n} n!}}\frac{{\sqrt \pi }}{2}.$$

Second approach. Writing $$\int_0^\infty {x^{2n} e^{ - x^2 } dx} = \frac{1}{2} \frac{{\sqrt {2\pi (1/2)} }}{{\sqrt {2\pi (1/2)} }} \int_{ - \infty }^\infty {x^{2n} \exp \bigg( - \frac{{x^2 }}{{2(1/2)}}\bigg)dx}$$ shows that $$\int_0^\infty {x^{2n} e^{ - x^2 } dx} = \frac{{\sqrt \pi }}{2}{\rm E}[X^{2n} ],$$ where ${\rm E}[X^{2n} ]$ is the $2n$-th moment of the Normal$(0,1/2)$ distribution. Now you can see how others, here for example, find ${\rm E}[X^{2n} ]$ (for a Normal$(0,\sigma^2)$ distribution). The simplest approach may be to use integration by parts. I'll leave it to you.

EDIT (in light of the OP's edit): Integration by parts gives the result, using induction, as follows: $$\int_0^\infty {x^{2(n + 1)} e^{ - x^2 } dx} = \frac{{2n + 1}}{2}\int_0^\infty {x^{2n} e^{ - x^2 } dx} = \frac{{[2(n + 1)]!}}{{2^{2(n + 1)} (n + 1)!}}\frac{{\sqrt \pi }}{2}.$$ For the base case $n=0$, note that $\int_0^\infty {e^{ - x^2 } dx} = \frac{{\sqrt \pi }}{2}$.

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Let's suppose that, one way or another, you know that $\displaystyle \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$. 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,

$$\int_{-\infty}^{\infty} e^{2tx - x^2} dx = \int_{-\infty}^{\infty} \left( \sum_{n \ge 0} \frac{2^n t^n x^n}{n!} \right) e^{-x^2} dx = \sum_{n \ge 0} \frac{2^n t^n}{n!} \int_{-\infty}^{\infty} x^n e^{-x^2} \, dx.$$

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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Do you mean the integral from 0 to infinity? –  Raeder Mar 15 '11 at 12:42
@Raeder: thank you for reminding me that I messed up the constant in the first statement, but no. The first step is messier if you take the integral from zero; you want the integral over the entire real line to get translational symmetry, but on the other hand $e^{-x^2}$ is even. –  Qiaochu Yuan Mar 15 '11 at 12:45

\textbf{Case 1:} For $n=0$. $\displaystyle \int^{\infty}_{0}{e^{-x^{2}}}dx=\frac{\sqrt{\pi}}{2}$.\ \ Now, we use the fact that $e^{-x^{2}}$ is an even function, and thus we have [ \displaystyle \int^{\infty}{0}{e^{-x^{2}}}dx=\frac{1}{2}\int^{+\infty}{-\infty}{e^{-x^{2}}}dx. ] Moreover, \begin{center} \begin{tabular}{lll} $\displaystyle\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx$ &=& $\displaystyle\sqrt{\left(\int^{+\infty}{-\infty}{e^{-x^{2}}}dx\right)\left(\int^{+\infty}{-\infty}{e^{-x^{2}}}dx\right)}$\ \ \ &=& $\displaystyle\sqrt{\left(\int^{+\infty}{-\infty}{e^{-y^{2}}}dy\right)\left(\int^{+\infty}{-\infty}{e^{-x^{2}}}dx\right)}$\ \ &=& $\displaystyle\sqrt{\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}{e^{-\left(x^{2}+y^{2}\right)}}dydx}$ \end{tabular} \end{center} Here, we use the fact that the variable in the integral is a dummy variable that is integrates out in the end and can be renamed from $x$ to $y$. Moreover, switching to polar coordinates then gives\ \begin{center} \begin{tabular}{lll} $\displaystyle\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx$ &=& $\displaystyle\sqrt{\int^{2\pi}{0}\int^{+\infty}{0}e^{-r^{2}r}drd\theta}$\ \ \ &=& $\displaystyle\sqrt{\int^{+\infty}{0}re^{-r^{2}}\int^{2\pi}{0}d\theta}$\ \ &=& $\displaystyle\sqrt{-\frac{1}{2}e^{-r^{2}}|^{\infty}{0}\cdot 2\pi}$\ \ &=& $\displaystyle\sqrt{\frac{1}{2}\cdot 2\pi}$\ \ &=& $\displaystyle\sqrt{\pi}$ \end{tabular} \end{center} And so $\displaystyle \int^{\infty}_{0}{e^{-x^{2}}}dx=\frac{1}{2}\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx=\frac{\sqrt{\pi}}{2}$.\ \ \textbf{Case 2:} For $n\geq 1$. Let [ a{n}=\displaystyle \int^{+\infty}{0}{x^{2n}e^{-x^{2}}}dx. ] Using integration by parts, let $u_{1}=x^{2n-1}$ so that $du_{1}=(2n-1)x^{2n-2}dx$, and $dv_{1}=xe^{-x^{2}}dx$ so that $\displaystyle v_{1}=-\frac{1}{2}e^{-x^{2}}$. Then \begin{center} \begin{tabular}{lll} $\displaystyle \int^{+\infty}_{0}{x^{2n}e^{-x^{2}}}dx$ &=& $\displaystyle -\frac{1}{2}e^{-x^{2}}x^{2n-1}|^{+\infty}{0}-\int^{+\infty}{0}{-\frac{1}{2}e^{-x^{2}}(2n-1)x^{2n-2}}dx$\ \ \ &=& $\displaystyle 0+ \frac{(2n-1)}{2}\int^{+\infty}{0}{x^{2n-2}e^{-x^{2}}}dx$\ \ \ &=& $\displaystyle \frac{(2n-1)}{2}\int^{+\infty}_{0}{x^{2n-2}e^{-x^{2}}}dx$ \end{tabular} \end{center} Using the integration by parts again, we let $u_{2}=x^{2n-3}$ so that $du_{2}=(2n-3)x^{2n-4}dx$, and let $dv_{2}=xe^{-x^{2}}dx$ so that $v_{2}=\displaystyle -\frac{1}{2}e^{-x^{2}}$. Again we have \begin{center} \begin{tabular}{lll} $\displaystyle \int^{+\infty}_{0}{x^{2n}e^{-x^{2}}}dx$ &=& $\displaystyle \frac{(2n-1)}{2}\int^{+\infty}{0}{x^{2n-2}e^{-x^{2}}}dx$\ \ \ &=& $\displaystyle \frac{(2n-1)}{2} \left(-\frac{1}{2}x^{2n-3}e^{-x^{2}}-\int^{+\infty}{0}{-\frac{1}{2}e^{-x^{2}}(2n-3)x^{2n-4}dx}\right)$\ \ \ &=& $\displaystyle \frac{(2n-1)}{2} \left(0-\int^{+\infty}{0}{-\frac{1}{2}e^{-x^{2}}(2n-3)x^{2n-4}dx}\right)$\ \ \ &=& $\displaystyle \frac{(2n-1)(2n-3)}{2^{2}} \int^{+\infty}_{0}{e^{-x^{2}}x^{2n-4}dx}$ \end{tabular} \end{center} Following the same process, we can obtain \begin{center} \begin{tabular}{lll} $a_{n}$ &=& $\displaystyle \int^{+\infty}{0}{x^{2n}e^{-x^{2}}}dx$\ \ \ &=& $\displaystyle\frac{(2n-1)(2n-3)(2n-5)(2n-7)\ldots (7)(5)(3)(1)}{2^{n}}\cdot a_{0}$\ \ \ &=& $\displaystyle\frac{(2n-1)(2n-3)(2n-5)\ldots(5)(3)(1)}{2^{n}}\cdot a_{0}\cdot\left(\frac{(2n-2)(2n-4)(2n-6)\ldots(6)(4)(2)}{(2n-2)(2n-4)(2n-6)\ldots(6)(4)(2)}\right)$\ \ \ &=& $\displaystyle\frac{(2n-1)(2n-3)(2n-5)\ldots(5)(3)(1)}{2^{n}}\cdot a_{0}\cdot \frac{(2n-2)(2n-4)(2n-6)\ldots(6)(4)(2)}{2^{n-1}(n-1)(n-2)(n-3)\ldots(3)(2)(1)}$\ \ \ &=& $\displaystyle \frac{(2n-1)(2n-2)(2n-3)\ldots (3)(2)(1)}{2^{n}2^{n-1}(n-1)!}\cdot a_{0}$\ \ \ &=& $\displaystyle \frac{(2n-1)!}{2^{2n-1}(n-1)!}\cdot a_{0}$\ \ \ &=& $\displaystyle \frac{(2n-1)!}{2^{2n-1}(n-1)!}\cdot\frac{2n}{2n}\cdot a_{0}$\ \ \ &=& $\displaystyle \frac{(2n)!}{2^{2n-1}2(n-1)!n}\cdot a_{0}$\ \ \ &=& $\displaystyle \frac{(2n)!}{2^{2n}(n)!}\cdot a_{0}$\ \ \ &=& $\displaystyle \frac{(2n)!}{2^{2n}(n)!}\cdot \frac{\sqrt{\pi}}{2}$ \ \ \ from \textbf{Case 1} \end{tabular} \end{center}

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The proof requires one line only (or less). Take a look at the formula $58$ here and guess on your own the way.

Chris.

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