Hi,

I'm trying to figure out how the integral representation of the Digamma function works.

I understand that the Digamma function is the logarithmic derivative of the Gamma function. Therefore

$\displaystyle \Psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$

Since $\displaystyle \Gamma(z)=\int^\infty_0e^{-t}t^{z-1}dt$

then I would think

$\displaystyle \Gamma'(z)=\int^\infty_0e^{-t}t^{z-1}\ln(t)dt$

Anyway, the integral expression for the Digamma function is given by

$\displaystyle \Psi(z)=\int^\infty_0\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}dt$

How would I derive this formula?

Should I instead look at another representation of the Gamma function and work with that?

Thank you.