# 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!

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

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,

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.

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 $\int_0^\infty {e^{ – x^2 } dx} = \frac{{\sqrt \pi }}{2}$.

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,

$$\int_{-\infty}^{\infty} x^{2n} e^{-x^2} \, dx = 2 \int_0^{\infty} x^{2n} e^{-x^2} \, dx.$$

Case 1: For $n=0$ we have $\int^{\infty}_{0}{e^{-x^{2}}}dx=\frac{\sqrt{\pi}}{2}$.
Now, we use the fact that $e^{-x^{2}}$ is an even function, and thus we have
$\int^{\infty}_{0}{e^{-x^{2}}}dx=\frac{1}{2}\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx.$
Moreover,
$$\int^{+\infty}_{-\infty}{e^{-x^2}}dx = \sqrt{\left(\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx\right)\left(\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx\right)}=$$
$$= \sqrt{\left(\int^{+\infty}_{-\infty}{e^{-y^{2}}}dy\right)\left(\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx\right)} = \sqrt{\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}{e^{-\left(x^{2}+y^{2}\right)}}dydx}$$

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
$$\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx =\sqrt{\int^{2\pi}_{0}\int^{+\infty}_{0}e^{-r^{2}r}drd\theta} =$$
$$=\sqrt{\int^{+\infty}_{0}re^{-r^{2}}\int^{2\pi}_{0}d\theta} = \sqrt{-\frac{1}{2}e^{-r^{2}}|^{\infty}_{0}\cdot 2\pi} =$$
$$= \sqrt{\frac{1}{2}\cdot 2\pi} = \sqrt{\pi}$$
And so $\int^{\infty}_{0}{e^{-x^{2}}}dx=\frac{1}{2}\int^{+\infty}_{-\infty}{e^{-x^{2}}}dx=\frac{\sqrt{\pi}}{2}$.

Case 2: For $n\geq 1$. Let $a_n= \int^{+\infty}_{0}{x^{2n}e^{-x^{2}}}dx$.
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 $v_{1}=-\frac{1}{2}e^{-x^{2}}$. Then
$$\int^{+\infty}_{0}{x^{2n}e^{-x^{2}}}dx = -\frac{1}{2}e^{-x^{2}}x^{2n-1}|^{+\infty}_{0}-\int^{+\infty}_{0}{-\frac{1}{2}e^{-x^{2}}(2n-1)x^{2n-2}}dx =$$
$$= 0+ \frac{(2n-1)}{2}\int^{+\infty}_{0}{x^{2n-2}e^{-x^{2}}}dx = \frac{(2n-1)}{2}\int^{+\infty}_{0}{x^{2n-2}e^{-x^{2}}}dx$$

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 $v_{2}= -\frac{1}{2}e^{-x^{2}}$. Again we have
$$\int^{+\infty}_{0}{x^{2n}e^{-x^{2}}}dx = \frac{(2n-1)}{2}\int^{+\infty}_{0}{x^{2n-2}e^{-x^{2}}}dx =$$
$$= \frac{(2n-1)}{2} \left(-\frac{1}{2}x^{2n-3}e^{-x^{2}}-\int^{+\infty}_{0}{-\frac{1}{2}e^{-x^{2}}(2n-3)x^{2n-4}dx}\right) = \frac{(2n-1)}{2} \left(0-\int^{+\infty}_{0}{-\frac{1}{2}e^{-x^{2}}(2n-3)x^{2n-4}dx}\right) =$$
$$= \frac{(2n-1)(2n-3)}{2^{2}} \int^{+\infty}_{0}{e^{-x^{2}}x^{2n-4}dx}$$
Following the same process, we can obtain
$$a_n = \int^{+\infty}_{0}{x^{2n}e^{-x^{2}}}dx = \frac{(2n-1)(2n-3)(2n-5)(2n-7)\ldots (7)(5)(3)(1)}{2^{n}}\cdot a_{0} =$$
$$=\frac{(2n-1)(2n-3)(2n-5)\ldots(5)(3)(1)}{2^{n}}\cdot a_{0}\cdot\left(\frac{(2n-2)(2n-4)(2n-6)\ldots(6)(4)(2)}{(2n-2)(2n-4)(2n-6)\ldots(6)(4)(2)}\right) = \frac{(2n-1)(2n-3)(2n-5)\ldots(5)(3)(1)}{2^{n}}\cdot a_{0}\cdot \frac{(2n-2)(2n-4)(2n-6)\ldots(6)(4)(2)}{2^{n-1}(n-1)(n-2)(n-3)\ldots(3)(2)(1)} =$$
$$= \frac{(2n-1)(2n-2)(2n-3)\ldots (3)(2)(1)}{2^{n}2^{n-1}(n-1)!}\cdot a_{0} = \frac{(2n-1)!}{2^{2n-1}(n-1)!}\cdot a_{0} =$$
$$= \frac{(2n-1)!}{2^{2n-1}(n-1)!}\cdot\frac{2n}{2n}\cdot a_{0} = \frac{(2n)!}{2^{2n-1}2(n-1)!n}\cdot a_{0} =$$
$$= \frac{(2n)!}{2^{2n}(n)!}\cdot a_{0} = \frac{(2n)!}{2^{2n}(n)!}\cdot \frac{\sqrt{\pi}}{2}$$
from Case 1.