# Find the residue of g(z) at z=-2 using Laurent Expansion

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1. Jan 13, 2015

1. The problem statement, all variables and given/known data
Find the residue at z=-2 for

$$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$

2. Relevant equations

$$\psi(-z)$$ represents the digamma function, $$\zeta(z)$$ represents the Riemann-Zeta-Function.

3. The attempt at a solution
I know that:

$$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$

Let $$z \to -1 - z$$ to get:

$$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$

Therefore we divide by the other part to get:

$$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$

I have to somehow get the coefficient of $$\frac{1}{z+2}$$ because I want to evaluate the residue at z=-2

The problem is I cant ever get a factor of $$\frac{1}{z+2}$$ because of the cubed (z+2) in the bottom. What should I do?

2. Jan 13, 2015

### RUber

The residue for a pole of order m at k is given by:
$lim _{z->k} [\frac{1}{(m-1)!} \frac{ d^{m-1}}{dz^{m-1}} ((z-k)^m f(z))]$
For this case, what is m? k = -2. Do you know the derivatives of $\psi(z)$?

3. Jan 13, 2015

### Ray Vickson

First: be very careful when using a series expansion such as
$$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$
This may have limited applicability (that is, may be valid only in a subset of $\mathbb{C}$) and may need to be extended to other regions by some standard techniques of "analytic continuation". Basically, you need to know whether or not $\psi(-z)$ has singularities, and where they are. For example, can it have a pole at $z = -2$? If so, that would change the residue.

So, your first order of business should be to understand fully the nature of the singularities of $g(z)$, at least near $z = -2$.