How to evaluate $\int_0^\infty m^{x+1}e^{-2m}dm$ as $\Gamma(x+2)\frac{1}{2}^{x+2}$?

$$\int_0^\infty \frac{m^{x+1}e^{-2m}}{\Gamma(x+1)\Gamma(2)}dm =\frac{\Gamma(x+2)\frac{1}{2}^{x+2}}{\Gamma(x+1)\Gamma(2)}$$

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.

• Make the substitution $z = 2m$. – guy Oct 10 '18 at 19:25
• @guy - may as well expand that into an answer, since no-one's going to do better... – jbowman Oct 10 '18 at 19:32
• Don't you mean t = 2m? – hippo Oct 10 '18 at 19:39
• @hippo same thing. call the dummy variable whatever you'd like. Although that might be a more judicious choice if you count the letter $z$ as having been used before – Taylor Oct 10 '18 at 20:02

Let $$z = 2m$$, then $$\frac{\text{d}z}{\text{d}m} = 2$$ and your integral equals $$\int_0^\infty m^{x+1}e^{-2m}\text{d}m = 2^{-x-2}\int_0^{\infty} z^{x+1} e^{-z}\text{d}z = 2^{-x-2} \Gamma(x+2).$$
Alternatively: $$\int_0^\infty m^{x+1}e^{-2m}\text{d}m = \Gamma(x+2) 2^{-(x+2)}\int_0^\infty \frac{1}{\Gamma(x+2) 2^{-(x+2)}} m^{x+1}e^{-2m}\text{d}m$$ and we can recognize the last integral is $$1$$ because it's the integral of a $$\text{Gamma}(x+2,2^{-1})$$ density.