How to integrate using differentiation under the integral sign

Question

$$\int _0^{\infty }x^2e^{-ax^2}dx$$

where $a > 0,$ given that $\displaystyle\int _0^{\infty }e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}} $. Not sure where to begin with this

Answer 1

Hint: For $a \ne 0$: $$I=\int _0^{\infty }x^2e^{-ax^2}dx=-\dfrac 1 {2a}\int _0^{\infty }x(-2axe^{-ax^2})dx$$ Integrate by parts.

Answer 2

For fixed $x$, let $f_x(a)=e^{-ax^2}$. Then $\frac d {da}f_x(a)=-x^2e^{-ax^2}$. So, given $$\int_0^{\infty}f_x(a)dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}$$ we see that $$\int_0^{\infty}x^2f_x(a)dx=\int_0^\infty -\frac d {da}f_x(a)dx=-\frac d {da}\frac{1}{2}\sqrt{\frac{\pi }{a}}=\frac{\sqrt\pi}{4\sqrt{a^3}}$$