# How to obtain this?

1. Apr 11, 2010

### S_David

Hello,

I am reading some material that using mathematics extensively, and I encountered with the following result:

$$\frac{N}{\overline{\gamma}}\,\int_0^{\infty}\gamma\,\left[1-\mbox{e}^{-\gamma/\overline{\gamma}}\right]^{N-1}\,\mbox{e}^{-\gamma/\overline{\gamma}}\,d\gamma=\,\overline{\gamma}\sum_{k=1}^N\frac{1}{k}$$

How did they get there? I tried to use the binomial expansion and assemble the exponentials, but the result was totally different. Any hint will be highly appreciated.

2. Apr 11, 2010

### CompuChip

Note that the integrand looks very much like a derivative of
$$\left( 1 - e^{-\gamma / \overline{\gamma}} \right)^N$$
with respect to either gamma or gamma-bar.
Yes, you are right. the term $$\left[1-\mbox{e}^{-\gamma/\overline{\gamma}}\right]^N$$ is the CDF of $$\gamma$$. In the integral it is intended to find the statistical average (the expected value) of $$\gamma$$ which needs the PDF of $$\gamma$$ which is the derivative of the CDF with respect to $$\gamma$$. But I am still stuck. Any further hint?