My problems #17

By definitiion of gamma function

$\displaystyle\int_0^{\infty} e^{-x} x^3 = \displaystyle\int_0^{\infty} e^{-x}x^{4-1}$

$= \Gamma {4} = \boxed{6}$

Without using gamma function.

$\begin{aligned} I(a) & = \int_0^\infty e^{-ax}x^3 \ dx \\ & = - \int_0^\infty \frac {\partial^3 e^{-ax}}{\partial a^3} \ dx \\ & = \frac {\partial^3}{\partial a^3} \int^0_\infty e^{-ax} \ dx \\ & = \frac {\partial^3}{\partial a^3} \left(-\frac 1a\right) \\ & = \frac 6a \\ \implies I(1) & = \int_0^\infty e^{-x}x^3 \ dx = \boxed{6} \end{aligned}$