# Residue of f(z) involving digamma function

## Homework Statement

Find the residue of:

$$f(z) = \frac{(\psi(-z) + \gamma)}{(z+1)(z+2)^3} \space \text{at} \space z=n$$

Where $n$ is every positive integer because those $n$ are the poles of $f(z)$

## The Attempt at a Solution

This is a simple pole, however:

$$\lim_{z \to n} \frac{(z-n)\psi(-z)}{(z+1)(z+2)^3}$$

But this gives 0, which is incorrect?

RUber
Homework Helper
For n to be a simple pole of ##\psi (-z) ## you will need to be able to express ##\psi (-z) = (z+n)g(z)## where g(z) is analytic at z=-n. Be careful with your signs.
Using the expansion ##\psi (n)+\gamma= \sum_{k=1}^n-1 \frac 1k ## shows that for every positive integer n there will be a simple pole at z=-n, since the sum will include k=0.

For n to be a simple pole of ψ(−z)\psi (-z) you will need to be able to express ψ(−z)=(z+n)g(z)\psi (-z) = (z+n)g(z) where g(z) is analytic at z=-n. Be careful with your signs.
Using the expansion ψ(n)+γ=∑nk=1−11k\psi (n)+\gamma= \sum_{k=1}^n-1 \frac 1k shows that for every positive integer n there will be a simple pole at z=-n, since the sum will include k=0.

Yes I see this, but what about higher order poles? consider:

Suppose we are given:

Find the residue at z=n of $$(\psi(-z) + \gamma)^2$$

Where $$\psi(-z)$$ means the digamma function.

$$(\psi(-z) +\gamma)^2 = \left( 1 + \frac{1}{2} + \frac{1}{z} + \frac{1}{3} + \frac{1}{z-1} + \frac{1}{4} + \frac{1}{z-2} + ... + \frac{1}{n} + \frac{1}{z-n} + ... \right)\left( 1 + \frac{1}{2} + \frac{1}{z} + \frac{1}{3} + \frac{1}{z-1} + \frac{1}{4} + \frac{1}{z-2} + ... + \frac{1}{n} + \frac{1}{z-n} + ... \right)$$

We see that we have:

$$\frac{1}{z-n} \cdot \left( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + \frac{1}{z-1} + \frac{1}{z-2} + ... + \frac{1}{z-n} \right)$$

The pure coefficient of $$\frac{1}{z-n}$$ is simply:

$$\left( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right) = H_n$$

Where $$H_n$$ means the Harmonic Number.

But then we also had the other $$\frac{1}{z-n}$$ in the other parentheses so twice this.

$$2 \left( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \right) = 2H_n$$

But does finding the coefficient work for higher order residues?

Thanks!