This is a solution I had come across for a general case.$$$$
We will use $$\Gamma(x)=\int_0^{\infty}t^{x-1}e^{-t}\,dt$$ $$\Gamma(\frac{1}{2}+n)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}$$ & we will first solve the general case $$I=\int_0^{\infty}x^me^{-kx^n}\,dx$$ take $y=kn^x$ $$\Rightarrow I=\frac{1}{n\cdot k^{\frac{m+1}{n}}}\int_0^{\infty}y^{\frac{m+1}{n}-1}e^{-y}\,dy$$ $$I=\frac{1}{n\cdot k^{\frac{m+1}{n}}}\Gamma(\frac{m+1}{n})$$
Could someone please explain how $y=kn^x$ transforms $$I=\int_0^{\infty}x^me^{-kx^n}\,dx$$ to $$\Rightarrow I=\frac{1}{n\cdot k^{\frac{m+1}{n}}}\int_0^{\infty}y^{\frac{m+1}{n}-1}e^{-y}\,dy$$ Thanks very much!
Set $y=kx^n$ we get $dy=nkx^{n-1}dx$ and we transform the integral by substituting $x=\left(\frac{y}{k}\right)^{\frac{1}{n}}$ and $dx=\frac{dy}{nkx^{n-1}}$. The bounds are still $0$ and $+\infty$