Let $I = \displaystyle{}\int_{0}^{\infty}e^{-x^2}x^4dx\rightarrow (1)$
let $x^{2} = t\implies x= \sqrt{t}$
$2xdx = dt\implies dx = \dfrac{dt}{2x} \implies dx = \dfrac{dt}{2\sqrt{t}} = \dfrac{t^{-1/2}\:dt}{2}$
Put the values in equation $(1),$ we get
$I = \displaystyle{}\int_{0}^{\infty}e^{-t}. t^{2}\bigg( \frac{t^{-1/2}}{2}\bigg)dt$
$I = \displaystyle{}\frac{1}{2}\int_{0}^{\infty}e^{-t} .t^{2}t^{-1/2}dt$
$I = \displaystyle{}\frac{1}{2}\int_{0}^{\infty}e^{-t}.t^{3/2}dt$
We know that $\Gamma(n) = \displaystyle{}\frac{1}{2}\int_{0}^{\infty}e^{-x}.x^{n-1}dx\:\:\:;n>0$
$I = \displaystyle{}\frac{1}{2}\bigg[\int_{0}^{\infty}e^{-t}.t^{(5/2-1)}dt\bigg]$
$\therefore I = \dfrac{1}{2}\:\Gamma \left(\frac{5}{2}\right)$
Standard Results:
- $\Gamma(1) = 1$
- $\Gamma(\frac{1}{2}) = \sqrt{\pi}$
- $\Gamma(n+1) = n\Gamma(n)$
- $\Gamma(n+1) = n!$
Now$,I = \dfrac{1}{2}\:\Gamma \left(\frac{5}{2}\right) = \dfrac{1}{2}\cdot\Gamma(\frac{3}{2} + 1) = \dfrac{1}{2}\cdot\frac{3}{2}\Gamma(\frac{3}{2}) = \dfrac{1}{2}\cdot\frac{3}{2}\Gamma(\frac{1}{2} + 1) = \dfrac{1}{2}\cdot \frac{3}{2}\cdot \frac{1}{2}\Gamma (\frac{1}{2}) = \frac{3}{8} \cdot \sqrt{\pi}$
So, the correct answer is $I = \dfrac{3\:\sqrt{\pi}}{8}$