Evaluating integral $\int_0^\infty x^4e^{-2a x^2}dx $ I was wondering how I can evaluate this integral
$$\int_0^\infty x^4e^{-2a x^2}dx  $$
I can't seem to find any specific formulas for this and the non-specific ones are confusing. An image is attached illustrating how this integral looks. Thanks.
Integral image
 A: Differentiate the known integral $I(b)=\int_0^\infty e^{-bx^2}dx = \frac12\sqrt{\frac\pi b}$ to evaluate
$$\int_0^\infty x^4e^{-2a x^2}dx 
= \frac{d^2I(b)}{db^2}\bigg|_{b=2a}=\frac{3}{32a^{5/2}}\sqrt{\frac\pi2}
$$
A: Let
$$J(a)=\int_0^\infty x^4e^{-2ax^2}dx,$$
assuming $\text{Re}(a)>0$. Then let $u=2ax^2$ so that
$$J(a)=\frac1{4a}\int_0^\infty\left(\frac{u}{2a}\right)^{3/2}e^{-u}du=\frac{1}{2(2a)^{5/2}}\int_0^\infty u^{5/2-1}e^{-u}du,$$
Which is
$$J(a)=\frac{\Gamma(\tfrac52)}{2(2a)^{5/2}}=\frac{3}{32}\sqrt{\frac{\pi}{2a^5}}$$
A: There are already two nice solutions using Gamma Function and differentiation. I am now going to share how to find the integral in general by a reduction formula.
$$
I_{n}:=\int_{0}^{\infty} x^{2 n} e^{-2 a x^{2}} d x
$$
Using integration by parts, we have
$$
I_{n}=\frac{1}{2 n+1}\left[e^{-2 a x^{2}} x^{2 n+1}\right]_{0}^{\infty}+\frac{4 a}{2 n+1} \int_{0}^{\infty} x^{2(n+1)} e^{-2 a x} d x
$$
So
$$
\begin{aligned}
I_{n+1} &=\frac{2 n+1}{4 a} \cdot I_{n} \\
&=\frac{2 n+1}{4 a} \cdot \frac{2 n-1}{4 a} \cdots\frac{1}{4 a} \int_{0}^{\infty} e^{-2 a x^{2}} d x\\&= \boxed{\frac{(2 n+1) ! !}{(2 \sqrt{a})^{2 n+3}} \sqrt{\frac{\pi}{2}}}
\end{aligned}
$$
In particular,
$$
\boxed{I_{2}=\frac{3}{32 a^{\frac{5}{2}}} \sqrt{\frac{\pi}{2}}}
$$
