Integration Evaluation (Digamma function)

[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 something different. Any hint will be highly appreciated.

Thanks in advance

 

[fresh_42]

With $g(\gamma)=\left[1-e^{-\gamma/\overline{\gamma}} \right]^N$ we have for large $N$
\begin{align*}
\dfrac{N}{\overline{\gamma}}\int_0^\infty \gamma \left[1-e^{-\gamma/\overline{\gamma}}\right]^{N-1}e^{-\gamma/\overline{\gamma}} & = \int_0^\infty \gamma \,g'(\gamma) d\gamma\\
&= \left[\gamma\,g(\gamma) - \int g(\gamma)\,d\gamma\right]_0^\infty \\
&=\left[ \gamma + \gamma \sum_{k=1}^N \binom{N}{k} \left(-e^{-\gamma/\overline{\gamma}} \right)^{k} - \gamma - \sum_{k=1}^N \binom{N}{k} \int \left(-e^{-\gamma/\overline{\gamma}} \right)^{k} \,d\gamma \right]_0^\infty \\
&=\sum_{k=1}^N \binom{N}{k} \left[ \left(\gamma +\dfrac{\overline{\gamma}}{k}\right) \left(-e^{-\gamma/\overline{\gamma}} \right)^{k} \right]_0^\infty \\
&= - \sum_{k=1}^N \binom{N}{k} \left(0+\dfrac{\overline{\gamma}}{k}\right) \left(-e^{-0 /\overline{\gamma}} \right)^{k} \\
&=\overline{\gamma} \cdot \left(- \sum_{k=1}^N (-1)^k \binom{N}{k}\dfrac{1}{k}\right) \\
&= \overline{\gamma} \cdot \left( \overline{\gamma} +\psi(N+1) \right)\\
&= \overline{\gamma} \cdot \left( \overline{\gamma} +\log N +\dfrac{1}{2N}- \dfrac{1}{12N^2}+\dfrac{1}{120N^4} +\mathcal{O}\left(\dfrac{1}{N^6}\right)\right)
\end{align*}
with the polygamma function $\psi$ and its series expansion at $N=\infty$. On the other hand we have
\begin{align*}
\overline{\gamma}\sum_{k=1}^N \dfrac{1}{k} &= \overline{\gamma} H_N = \overline{\gamma} \left(\log N + \overline{\gamma} +\dfrac{1}{2N}- \dfrac{1}{12N^2}+\dfrac{1}{120N^4} +\mathcal{O}\left(\dfrac{1}{N^6} \right) \right)
\end{align*}
See also http://fractional-calculus.com/gamma_digamma.pdf