How do I show the following?
$$\int_{0}^{\infty} x^n e^{-ax}\,dx = \frac{n}{a} \int_{0}^{\infty} x^{n-1}\,e^{-ax}\,dx. $$
I tried integration by parts, but it gets one big mess. I do seem to get $n/a$ somewhere, though. However, I'm mostly struggling with the fact that I have to evaluate the expression $-x^n e^{-ax}/a$ for 0 to infinity, because I would get infinity over infinity and I don't know how to apply l'Hôspital's rule with the number $e$.
I'll appreciate all input.
Integration by parts is the easiest way to get to this. It seems like you have an issue with computing
$ \lim\limits_{x \to \infty} \frac{x^n}{e^{ax}} $. There's a number of ways to do this, but because you're familiar with L'Hôpital's rule, let's go that way. Note that we assume $a>0$ below.
\begin{align} \lim_{x \to \infty} \frac{x^n}{e^{ax}} &= \lim_{x \to \infty} \frac{nx^{n-1}}{ae^{ax}} ~~~~\left(\frac{\infty}{\infty} \text{form} \right) \\ &= \lim_{x \to \infty} \frac{n(n-1)x^{n-2}}{a^2e^{ax}} ~~~~\left(\frac{\infty}{\infty} \text{form} \right) \\ &\text{$$} ~~~~~~~~~~~~~~~~~~~\vdots \\ &= \lim_{x \to \infty} \frac{\frac{n!}{1!} x}{a^{n-1}e^{ax}} ~~~~\left(\frac{\infty}{\infty} \text{form} \right) \\ &= \lim_{x \to \infty} \frac{n!}{a^n e^{ax}} ~~~~\left(\frac{\text{constant}}{\infty} \text{form} \right) \\ &= 0. \end{align}
The usual argument runs along the lines of $e^{-ax}$ goes to zero faster than any polynomial in $x$, so $P(x) e^{-ax} \to 0$ as $x \to \infty$ for all polynomials. This is saying the same thing as above but without having to write all that out.
Finally,
\begin{align} \int_0^\infty x^n e^{-ax} \mathrm{d}x &= \lim_{b \to \infty} \left( -\frac{1}{a}b^n e^{-ab} + 0 \right) - \int_0^\infty n x^{n-1} \cdot -\frac{e^{-ax}}{a} \mathrm{d}x \\ &= 0 + \frac{n}{a} \int_0^\infty x^{n-1} e^{-ax} \mathrm{d} x \end{align}
Voila!
