$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$ How to evaluate this integral? 
$$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$$
I found similar problem, but don't know how to apply them here.
What do I have to substitute?
 A: If we replace $x$ with $\sqrt{\frac{y}{2}}$ we have:
$$ I = \frac{1}{4\sqrt{2}}\int_{0}^{+\infty}y^{1/2}e^{-y}\,dy = \frac{\Gamma\left(\frac{3}{2}\right)}{4\sqrt{2}}=\color{red}{\frac{1}{8}\sqrt{\frac{\pi}{2}}}.$$
We have:
$$\Gamma\left(\frac{3}{2}\right)=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$$
since $\Gamma(x+1)=x\cdot \Gamma(x)$ follows from integration by parts and $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ since you know in advance the value of $\int_{-\infty}^{+\infty}e^{-x^2}\,dx$, i.e. $\sqrt{\pi}$.
A: Putting
$$\begin{align}&u=x&u'=1\\{}\\&v'=xe^{-2x^2}&v=-\frac14e^{-2x^2}\end{align}$$
we get your integral is
$$I=\overbrace{\left.-\frac14xe^{-2x^2}\right|_0^\infty}^{=0}+\frac14\int_0^\infty e^{-2x^2}dx$$
Now substitute $\;u=\sqrt2\,x\;\implies\;du=\sqrt2\,dx\;$ , and we get
$$I=\frac1{4\sqrt2}\int_0^\infty e^{-u^2}du=\frac{\sqrt\pi}{8\sqrt2}$$
A: We define the function $I(a)$ as
$$\begin{align}
I(a)&\equiv\int_0^{\infty}e^{-ax^2}dx\\\\
&=\frac{1}{2}\sqrt{\frac{\pi}{a}} \tag 1
\end{align}$$
Now, taking a derivative with respect to $a$ in $(1)$ reveals that
$$\begin{align}
I'(a)&=-\int_0^{\infty}x^2e^{-ax^2}dx\\\\
&=-\frac14\sqrt{\pi}a^{-3/2} \tag 2
\end{align}$$
Setting $a=2$ in $(2)$ and multiplying by $-1$ gives the desired result

$$\begin{align}
-I'(2)&=\int_0^{\infty}x^2e^{-2x^2}dx\\\\
&=\frac18\sqrt{\frac{\pi}{2}}
\end{align}$$


NOTE:
We can extend the usefulness of this approach by noting that 
$$\begin{align}
I^{(n)}(a)&\equiv \frac{d^n}{dx^n}\int_0^{\infty}e^{-ax^2}dx\\\\
&=(-1)^n\int_0^{\infty}x^{2n}e^{-ax^2}dx\\\\
&=(-1)^n\frac{\sqrt{\pi}}{2}\frac{(2n-1)!!}{2^n}a^{-(2n+1)/2}\\\\
\int_0^{\infty}x^{2n}e^{-ax^2}dx&=\frac{\sqrt{\pi}}{2}\frac{(2n-1)!!}{2^n}a^{-(2n+1)/2}
\end{align}$$
We can use $(2n-1)!!=\frac{(2n)!}{2^n\,n!}$ to write
$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}x^{2n}e^{-ax^2}dx=\frac{\sqrt{\pi}}{2}\frac{(2n)!}{4^n\,n!}a^{-(2n+1)/2}}$$
A: There are often several ways to evaluate some integrals. Here is one of the less standard ways. 
Consider the integral
\begin{align}
I(a) = \int_{0}^{\infty} x^2 \, e^{- a x^2} \, dx
\end{align}
for which 
\begin{align}
I(a) &= - \partial_{a} \, \int_{0}^{\infty} e^{-a x^2} \, dx 
= - \frac{1}{2} \, \partial_{a} \left( \sqrt{\frac{\pi}{a}} \right) 
= \frac{1}{4a} \, \sqrt{\frac{\pi}{a}}. 
\end{align}
Where $a=2$ the result desired is obtained. 
Integral evaluation for clarity:
\begin{align}
J(a) &= \int_{0}^{\infty} e^{-a x^{2}} \, dx \\
&= \frac{1}{2 \sqrt{a}} \, \int_{0}^{\infty} e^{-t} \, t^{-1/2} \, dt \hspace{10mm} t= a x^{2} \\
&= \frac{1}{2\sqrt{a}} \, \sqrt{\frac{\pi}{a}}.  
\end{align}
Further calculations:
\begin{align}
\int_{0}^{\infty} x^{2n} \, e^{-a x^{2}} \, dx &= (-1)^{n} \partial_{a}^{n} \left( \frac{1}{2} \sqrt{\frac{\pi}{a}} \right) \\
&= \frac{(2n)!}{2^{2n+1} \, n! \, a^{n}} \, \sqrt{\frac{\pi}{a}} 
\end{align}
