For the digamma function defined by \begin{equation} \psi^0(z)=\frac{\mathrm d}{\mathrm d z}\ln \Gamma(z);\quad z\in \mathbb{C}=Z_\Re+ i Z_\Im. \end{equation} where $\Gamma(z)=\int_0^\infty \mathrm d x e^{-x} x^{z-1}$ is the Gamma function.
If there a way to write down an closed form expression for $\Re \psi^0(z) $ and $\Im \psi^0(z) $?
Looking at the integral representations like \begin{equation} \psi^0(z)=\int_0^{\infty } \left(\frac{e^{-t}}{t}-\frac{e^{t (-z)}}{1-e^{-t}}\right) \, dt \end{equation} and others it looks like I can't separate out the real and imaginary part.
Are there any other series representation where this is possible? Or are there some other tricks that I can use?
I may be missing something, but can you not write
$e^{t (-z)}= e^{-t Z_\Re}\cos (t \: Z_\Im) - i \: e^{-t Z_\Re} \sin(t \: Z_\Im )$ ?
Hence $$\psi^0(z)=\int_0^{\infty} \left(\frac{e^{-t}}{t}-\frac{e^{-t Z_\Re}\cos (t \: Z_\Im)}{1-e^{-t}}\right) dt $$ $$ + \, i \int_0^{\infty} \left( \frac{\: e^{-t Z_\Re} \sin(t \: Z_\Im )}{1-e^{-t}}\right)dt $$
This gives a closed integral expressions for the real and imaginary parts or were you looking for another form for the separation?