If we have a function of the form $f(z)=\Gamma(z)\zeta(z)$ where $\Gamma(z)$ is the standard gamma function, and $\zeta(z)$ is the Riemann zeta function, then how would we calculate the residues?
I know that the Riemann zeta function has a pole at $z=1$ with a residue of 1, and that the gamma function has residues, at the poles $z=-n$ for $n\in\mathbb{Z}^+$, given by
$$\text{Res}_{z=-n}=\frac{(-1)^n}{n!},$$
but how would we find the residues of the product of the function?
Best Answer
We can always determine the residue of a product by multiplying the Laurent expansions of the factors. Generally, that is tedious (and doesn't lead to a closed form expression), but in the case where one of the factors has a simple pole, and the other factor is holomorphic at the point in question, the situation is quite simple: If
\begin{align} f(z) &= \sum_{n = -1}^{+\infty} a_n \cdot (z-z_0)^n,\\ g(z) &= \sum_{k = 0}^{+\infty} b_k\cdot (z-z_0)^k, \end{align}
then it is easy to see that the coefficient of $(z-z_0)^{-1}$ in the product is $a_{-1}\cdot b_0$.
Thus in this case, we have the simple
$$\operatorname{Res} (f\cdot g; z_0) = g(z_0) \cdot \operatorname{Res}(f; z_0).\tag{$\ast$}$$
For the product of the $\Gamma$-function and the Riemann $\zeta$-function, we are always in that situation. All poles of either factor are simple, and the two functions have no common pole. So $\operatorname{Res} (\Gamma\cdot \zeta; 1) = \Gamma(1)\cdot \operatorname{Res}(\zeta;1) = 1\cdot 1 = 1$, and for $n \in \mathbb{N}$ (where we use the convention $0\in \mathbb{N}$), $\Gamma$ has a simple pole at $-n$, and we find
$$\operatorname{Res}(\Gamma\cdot\zeta; -n) = \zeta(-n)\operatorname{Res}(\Gamma; -n).$$