$\int_0^\infty \frac{m^{x+1}e^{-2m}}{\Gamma(x+1)\Gamma(2)}dm =\frac{\Gamma(x+2)\frac{1}{2}^{x+2}}{\Gamma(x+1)\Gamma(2)}$
How does the left side equal the right side? I understand that the gamma function is $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$
and that $\Gamma(x+2) = \int_0^\infty m^{x+2-1} e^{-m} dm$
However I am missing something to understand where the $\frac{1}{2}^{x+2}$ comes from.