# integrate $I = \int_0^\infty \frac{m^m\gamma^{m-1}e^{\frac{-m\gamma}{\bar{\gamma}}}}{\Gamma(m)\bar{\gamma}^m} d\gamma$

I have to find the following integral which is basically a Nakagami-m distribution:

$$I = \int_0^\infty \frac{m^m\gamma^{m-1}e^{\frac{-m\gamma}{\bar{\gamma}}}}{\Gamma(m)\bar{\gamma}^m} d\gamma$$

Please note that $m$ is an integer and $\Gamma$ represents the gamma function.

I have computed it using the Laplace transform and the answer comes out to be 1. But I am not sure because when I simulate my equation (where $I$ is used) I am not gettting the right result.

Any help/hints will be highly appreciated.

Cordially,

• What is $\bar{\gamma}$? May 23 '14 at 13:26
• @Daniel Fischer: $\bar \gamma$ is the mean value of $\gamma$ and is a constant. May 23 '14 at 13:29

Indeed, $I = 1$. We can take the constant $\frac{1}{\Gamma(m)}$ out of the integral and group things a little, to get

$$I = \frac{1}{\Gamma(m)}\int_0^\infty \left(\frac{m\gamma}{\bar{\gamma}}\right)^{m-1}e^{-\frac{m\gamma}{\bar{\gamma}}}\,\frac{m}{\bar{\gamma}}\,d\gamma.$$

The substitution $x = \frac{m\gamma}{\bar{\gamma}}$ then brings it into the form

$$I = \frac{1}{\Gamma(m)}\int_0^\infty x^{m-1}e^{-x}\,dx,$$

and the last integral is easily recognised as $\Gamma(m)$.

• Thank you very much @Daniel Fischer May 23 '14 at 13:56