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

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.

  • 4
    $\begingroup$ Make the substitution $z = 2m$. $\endgroup$ – guy Oct 10 '18 at 19:25
  • $\begingroup$ @guy - may as well expand that into an answer, since no-one's going to do better... $\endgroup$ – jbowman Oct 10 '18 at 19:32
  • $\begingroup$ Don't you mean t = 2m? $\endgroup$ – hippo Oct 10 '18 at 19:39
  • $\begingroup$ @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 $\endgroup$ – 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.

Sorry, guy :)


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