Sum of Residues of $\psi^2(-z)$

Question

Compute the Sum of residues of $f(z) = \psi^2(-z)$, where $\psi(z)$ is the digamma function.

There are singularities for $z= 1, 2, 3, \ldots$, i.e. for all natural numbers.

But how do I compute the sum of the residues? It will be an infinite sum. but how should I do it?

Thanks. Also what order are the singularities (poles)?

I believe it has something to do with Laurent series, but I am not sure...

Answer 1

The Laurent expansion around negative integers is $$\psi(-z) +\gamma=\frac{1}{z-n}+H_n +\sum_{k=1}^\infty(-1)^{k} (H_n^{(k+1)}-\zeta(k+1))(z-n)^k $$

By squaring

$$\psi^2(-z) =\left(\frac{1}{z-n}+H_n-\gamma +\sum_{k=1}^\infty(-1)^{k} (H_n^{(k+1)}-\zeta(k+1))(z-n)^k\right)^2 $$

Hence the residue is $2(H_n-\gamma)$.