Expected Value

$\displaystyle \displaystyle\text{E}[X]=\frac{1}{\Gamma\left(\frac{2}{5}\right)20^{2/5}}\lim_{b\to\infty}\int_0^bx^{2/5}e^{-x/20}dx$

$\displaystyle \displaystyle\Gamma\left(\frac{2}{5}\right)=\lim_{ b\to\infty}\int_0^bx^{-3/5}e^{-x}dx$

How do I solve this Gamma function?