Integrate $\int_0^{\infty} r^4e^{-r} dr$ How is it possible to integrate this without having to use integration by parts several times? I can't see a substitution that would work.
 A: Differentiate the equation $\int_0^\infty e^{-tr}\, dr=1/t$ with respect to $t$, and then use induction. You will find that $\int_0^\infty r^n e^{-tr}\, dr=n!\,t^{-n-1}$. Then set $t=1$.
(This tip is from Folland's real analysis book)
A: If you know the Euler gamma function, you have
$$
\Gamma (s)=\int_0^{+\infty} r^{s-1}e^{-r} dr
$$ giving
$$
\int_0^{+\infty} r^{4}e^{-r} dr=\Gamma (5)=4!=24.
$$
A: You may use the prepared tables for calculation of Laplace's Transform. However, if you want to analytically prove the integration, you may have to use the method of integration by parts or other analytical methods. You can find ready tables for Laplace's Transform by searching the net.
$$\begin{array}{l}L\left\{ {f\left( t \right)} \right\} = F\left( s \right) = \int\limits_0^\infty  {f\left( t \right){e^{ - st}}dt} \\L\left\{ {{t^p}} \right\} = F\left( s \right) = \int\limits_0^\infty  {{t^p}{e^{ - st}}dt} ;p >  - 1\\\int\limits_0^\infty  {{t^p}{e^{ - st}}dt}  = \frac{{\Gamma \left( {p + 1} \right)}}{{{s^{p + 1}}}}\\L\left\{ {{t^4}} \right\} = F\left( 1 \right) = \int\limits_0^\infty  {{t^4}{e^{ - t}}dt} ;p = 4,s = 1\\\int\limits_0^\infty  {{t^4}{e^{ - t}}dt}  = \frac{{\Gamma \left( {4 + 1} \right)}}{{{1^{4 + 1}}}} = \Gamma \left( 5 \right) = 4! = 24\end{array}$$
