resolving integral using gamma function

Question

My book solves this integral \begin{equation} \int_0^{\infty} y^3e^{-ay} dy \end{equation} using gamma function as \begin{equation} \frac{1}{a^4}\Gamma(4) \end{equation} why is this true?

Answer 1

Hint

The change of variable $t=ay$ is useful.

Edit

We have \begin{equation} \int_0^{\infty} y^3e^{-ay} dy=\int_0^\infty \left(\frac t a\right)^3e^{-t}\frac{dt}{a}=\frac 1 {a^4}\int_0^\infty t^3e^{-t}dt=\frac{\Gamma(4)}{a^4} \end{equation}

Answer 2

Besides to Sami's post , note that: $$\Gamma(n+1)=\int_0^{\infty}x^n\exp(-x)dx=\lim_{M\to\infty}\int_0^Mx^n\exp(-x)dx=\cdots=n\Gamma(n),~~n>0$$