Question

Prove the integral:

\[\large \int_{0}^{\infty}x^ne^{-ax}dx=\frac{n!}{a^{n+1}} \] where \(n \ge 0, a>0\)

Answer 1

Integrate by parts, letting
\[\large\begin{matrix}
u=x^n&&&dv=e^{-ax}~dx\\
du=nx^{n-1}~dx&&&v=-\frac{1}{a}e^{-ax}
\end{matrix}\]
\[\large\int_0^\infty x^ne^{-ax}~dx=\color{red}{-\frac{1}{a}\bigg[x^ne^{-ax}\bigg]_0^\infty}+\frac{n}{a}\int_0^\infty x^{n-1}e^{-ax}~dx\]
The red term (the \(uv\) part of the IBP formula) will always be zero for \(n>0\), so really you will reduce this integral to something like this:
\[\large\begin{align*}\int_0^\infty x^ne^{-ax}~dx&=\cdots=\frac{n(n-1)(n-2)\cdots(2)(1)}{a\cdot a\cdot a\cdots a\cdot a}\int_0^\infty e^{-ax}~dx\\
&=\frac{n!}{a^n}\int_0^\infty e^{-ax}~dx\end{align*}\]

Answer 2

Ah thank you!!