\(\displaystyle = \int_0^{\infty} \frac{ e^{-zx} }{ 1- e^{-x} }~dx \)

(Happy)

Thanks! I think I just overlooked the substitution process. I completely understand it now though.

Anyway, now I understand the integral representation for \(\displaystyle \Psi(z)\)

However, I have one more question (and this question is open to anyone). I want to see how

\(\displaystyle \Psi(z)=H_{n-1}-\gamma\)

Where \(\displaystyle H_{n}\) are the Harmonic numbers and \(\displaystyle \gamma\) is the Euler-Mascheroni constant.

I want to derive this relationship using the integral representation of the Digamma function. I tried doing substitution, but I'm stuck. Here is my work so far:

For

\(\displaystyle \Psi(x)=\int_0^{\infty} [\frac{e^{-t}}{t} - \frac{ e^{-xt}}{ 1 - e^{-t}}]~dt\)

Let \(\displaystyle u=e^{-t}\), therefore \(\displaystyle -\ln(u)=t\) and \(\displaystyle -\frac{1}{u}du=dt\)

Substituting, I get

\(\displaystyle \int^0_1 [\frac{u}{-\ln(u)}-\frac{e^{-x\ln(u)}}{1-u}][-\frac{1}{u}]du\)

which simplifies to

\(\displaystyle -\int^1_0 [\frac{1}{\ln(u)}+\frac{u^{-x}u^{-1}}{1-u}]du\)

I know that \(\displaystyle \gamma=\int^1_0 [\frac{1}{\ln(u)}+\frac{1}{1-u}du\)

However, I'm not sure what to do from here. I also know the integral representation for the Harmonic numbers is given by:

\(\displaystyle H_n=\int^1_0 \frac{1-x^n}{1-x}dx\), but I don't see that occurring in the integral.

Of course, my substitution might not be the right choice. The main thing to do is to get from the Digamma integral representation into a form in which the Harmonic integral representation and an integral representing \(\displaystyle \gamma\) arises.

Thank you.(Happy)