Improper integral of
$$\int_0^\infty x^ne^{-x/a} \, dx$$
I tried to do it by using Partial Integral and looked for the trend of $a$ and $n$ but it's confusing and I cannot finish it
Is there any other method to solve this? (I think of the Laplace Transform, but cannot find any relation)
On
$$ \int_0^\infty x^ne^{-x/a} \, dx = a^{n+1} \int_0^\infty \left( \frac x a \right)^n e^{-x/a} \, \frac{dx} a = a^{n+1} \int_0^\infty u^n e^{-u} \, du = a^{n+1} \Gamma(n+1). $$
If you know that $\displaystyle \int_0^\infty u^n e^{-u} \,du = n!$ for non-negative integer $n,$ then you can use that here. You can show that by integrating by parts and getting $\displaystyle \int_0^\infty u^n e^{-u} \, du = n \int_0^\infty u^{n-1} e^{-u} \, du.$ Iterate that to get the result. In the process, you'll see $$ \int_0^\infty u^n e^{-u} \, du = \Big[ \cdots\cdots \Big]_0^\infty + n \int_0^\infty u^{n-1} e^{-u} \, du $$ and you can use L'Hopital's rule to show that that first term is $0.$
\begin{align} \int_0^{\infty} x^n e^{-\frac{x}{a}}\,dx &\overset{u=x/a}{=} \int_0^{\infty} (au)^ne^{-u}\,(a\,du)\\ &= a^{n+1}\int_0^{\infty}u^{(n+1)-1}e^{-u}\,du \\ &= a^{n+1}\Gamma(n+1) = a^{n+1}n! \end{align} if $n$ is an integer.