What is the method to obtain this integral result?

[EngWiPy]

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.

Thanks in advance

 

[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.
Maybe you can use that to your advantage.

 

[EngWiPy]

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.
Maybe you can use that to your advantage.

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?

Thanks in advance