Messy Gaussian Integral I am trying to understand how to better perform the following integral.
$$\int^{\infty}_{0} x^4 e^{\frac{-x^2}{\beta^2}}\mathrm{d}x$$
I've done a little research and found that $e^{-x^2}$ doesn't integrate easily, for it is the Gaussian integral. Many sources are pointing me to use polar coordinates. Is that the best way to go about this? How can I solve this integral?
 A: $$\int_{0}^{\infty} x^p e^{-ax}=\frac{\Gamma(p+1)}{a^{p+1}}.$$
So $$I=\int_{0}^{\infty}  x^4 e^{-x^2/b^2} dx=\int_{0}^{\infty} \frac{b^5}{2} t^{3/2} e^{-t} dt= \frac{b^5}{2}\Gamma(5/2)=\frac{3 \sqrt{\pi} b^5}{8}. $$
A: For $\lambda>0$ set
$$
I(\lambda) 
= \int_0^\infty e^{-\lambda x^2/\beta^2} dx
= \left\{ x=\frac{\beta}{\sqrt{\lambda}}y \right\}
= \int_0^\infty e^{-y^2} dy
= \frac{\beta}{\sqrt{\lambda}} \int_0^\infty e^{-y^2} dy
= \frac{\beta}{\sqrt{\lambda}} \frac{\sqrt{\pi}}{2}
.
$$
The last identity comes from the well-known result
$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$
and that $e^{-x^2}$ is even.
Taking the derivative gives
$$
I'(\lambda)
= \int_0^\infty \frac{-x^2}{\beta^2} e^{-\lambda x^2/\beta^2} dx
= \left( \frac{\beta}{\sqrt{\lambda}} \frac{\sqrt{\pi}}{2} \right)'
= \frac{-\beta}{2\lambda^{3/2}} \frac{\sqrt{\pi}}{2}
$$
and
$$
I''(\lambda)
= \int_0^\infty \left(\frac{-x^2}{\beta^2}\right)^2 e^{-\lambda x^2/\beta^2} dx
= \left(\frac{-\beta}{2\lambda^{3/2}} \frac{\sqrt{\pi}}{2}\right)'
= \frac{3\beta}{4\lambda^{5/2}} \frac{\sqrt{\pi}}{2}
.
$$
Taking $\lambda=1$ in the last equality gives
$$
I''(1) 
= \int_0^\infty \frac{x^4}{\beta^4} e^{-x^2/\beta^2} dx
= \frac{3\beta}{4} \frac{\sqrt{\pi}}{2}
$$
from which we conclude
$$
\int_0^\infty x^4 e^{-x^2/\beta^2} dx
= \frac{3\beta^5}{4} \frac{\sqrt{\pi}}{2}
= \frac{3\beta^5\sqrt{\pi}}{8}
.
$$
A: $$I_n=\int^{\infty}_{0} x^n e^{-\frac{x^2}{\beta^2}}dx$$ Let $x=\beta t$ to make
$$I_n=\beta^n \int^{\infty}_{0} t^n e^{-t^2}\,dt$$ If you computed the very first ones (integration by parts), the pattern is quite clear (at least to me) and
$$I_n=\frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)\beta^n$$
A: A couple of integration by parts can get you to the Gaussian integral, which has many resources online for derivations:
\begin{align*}
I=\int^{\infty}_{0} x^3 \left(x e^{-\frac{-x^2}{\beta^2}} \right)\; \mathrm{d}x &=x^3 \left(-\frac{\beta^2}{2} e^{-\frac{-x^2}{\beta^2}} \right) \bigg \rvert_0^{\infty}+\frac{3\beta^2}{2} \int_0^{\infty} x \left( xe^{-\frac{-x^2}{\beta^2}} \right) \\
&= -\frac{3\beta^4}{4} x \left( e^{-\frac{-x^2}{\beta^2}} \right) \bigg \rvert_0^{\infty} + \frac{3\beta^4}{4} \underbrace{\int_0^{\infty} e^{-\frac{-x^2}{\beta^2}} \; \mathrm{d}x}_{x/\beta \to t}\\
&=\frac{3\beta^5}{4} \underbrace{\int_0^{\infty} e^{-t^2} \; \mathrm{d}t}_{\sqrt{\pi}/2}\\
&= \frac{3\sqrt{\pi}\beta^5}{8}
\end{align*}
A: I just wanted to expand on Claude's answer. First a transformation:
$$\int_0^\infty x^n\exp\left(\frac{-x^2}{a^2}\right)\mathrm{d}x=a^{n+1}\int_0^\infty x^n\exp(-x^2)\mathrm{d}x$$
Let
$$I_n=\int_0^\infty x^n\exp(-x^2)\mathrm{d}x$$
Integration by parts. Let $u=\exp(-x^2)$, $\mathrm{d}u=-2x\exp(-x^2)\mathrm{d}x$, $\mathrm{d}v=x^n\mathrm{d}x$, $v=\frac{x^{n+1}}{n+1}$.
$$I_n=\int_0^\infty u~\mathrm{d}v=(uv)\big|^\infty_0-\int_0^\infty v~\mathrm{d}u$$
$$=\left(\frac{x^{n+1}}{n+1}\exp(-x^2)\right)\bigg|^\infty_0-\int_0^\infty-2x\exp(-x^2)\frac{x^{n+1}}{n+1}\mathrm{d}x$$
$$I_n=\frac{2}{n+1}I_{n+2}\implies I_{n+2}=\frac{n+1}{2}I_n$$
Now we need to compute $I_0,I_1$. It's obvious that $I_0=\sqrt{\pi}/2$.
$$I_1=\int_0^\infty x\exp(-x^2)\mathrm{d}x$$
Via a substitution $t=x^2$, $\mathrm{d}t=2x\mathrm{d}x$,
$$I_1=\frac{1}{2}\int_0^\infty e^{-t}\mathrm{d}t=\frac{1}{2}\Gamma(1)=\frac{1}{2}.$$
So
$$I_2=\frac{\sqrt{\pi}}{4}~;~I_3=\frac{1}{2}~;~I_4=\frac{3\sqrt{\pi}}{8},...$$
Since the recurrence relation jumps by two, we can separate the even and odd cases. For odd $n$,
$$I_n=I_1\cdot\left(\frac{(1+1)}{2}\frac{(3+1)}{2}\frac{(5+1)}{2}...\frac{n-2+1}{2}\right)=\frac{1}{2}\left(1\cdot 2\cdot 3\cdot~~...~~\cdot \frac{n-1}{2}\right)=\frac{1}{2}\left(\frac{n-1}{2}\right)!$$
And, since for $n\in\Bbb{N},~n!=\Gamma(n+1)$,
$$I_n=\frac{1}{2}\Gamma\left(\frac{n+1}{2}\right)$$
For even $n$, it's a slightly trickier.
$$I_n=I_0\left(\frac{(0+1)}{2}\frac{(2+1)}{2}\frac{(4+1)}{2}...\frac{n-2+1}{2}\right)=\frac{\sqrt{\pi}}{2}\left(\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot~~...~~\cdot\frac{n-1}{2}\right)$$
However one might notice that $\sqrt{\pi}=\Gamma(1/2)$. Using the recursive properties of the Gamma,
$$I_2=\frac{1}{2}I_0=\frac{1}{2}\frac{\Gamma(1/2)}{2}=\frac{\Gamma(3/2)}{2}$$
$$I_4=\frac{3}{2}I_2=\frac{3}{2}\frac{\Gamma(3/2)}{2}=\frac{\Gamma(5/2)}{2}$$
So it's easy to see in general that this actually lines up with what we got with the odd case.
$$I_n=\frac{1}{2}\Gamma\left(\frac{n+1}{2}\right)$$
Finally, $$\int_0^\infty x^n\exp\left(\frac{-x^2}{a^2}\right)\mathrm{d}x=a^{n+1}I_n=\frac{a^{n+1}}{2}\Gamma\left(\frac{n+1}{2}\right)$$
So the integral in question is
$$\frac{a^{4+1}}{2}I_4=\frac{a^5}{2}\Gamma(5/2)=\frac{3a^5\sqrt{\pi}}{8}.$$
