Let $x\in\mathbb{R},\,x>1$, and
$$f(s)= \frac{\zeta(2s)}{\zeta(s)}\frac{x^{s-1/2}}{s-\frac 12}$$
where $s=\sigma+it\in\mathbb{C}$. How do you calculate the residues of $f(s)$ in the critical strip (where $0<\sigma<1)$, assuming the Riemann hypothesis and that all poles are simple to begin with. So for example, how do you calculate the residue of $f(s)$ at $s=\rho=\frac 12+i\gamma$, where $\zeta(s)=0$? What happens without one or both assumptions?
If I use Mathematica for this, I get the expressions ahead, but please note that I would like to know is how Mathematica arrives at these expressions, and under what assumptions.
So, if $s=\frac 12$ Mathematica produces
$$\frac{\log x}{2\zeta\left(\frac{1}{2}\right)}-\frac{\zeta’\left (\frac{1}{2}\right)}{2\zeta\left (\frac{1}{2}\right)^2}+\frac{\gamma}{\zeta\left (\frac{1}{2}\right)}$$
And if $s=\rho=\frac 12+i\gamma$ and $\zeta(s)=0$ one gets
$$\frac{\zeta(2\rho)x^{i\gamma}}{i\gamma\zeta’(\rho)}$$
So again, my question is, how does Mathematica get to this result?
Best Answer
Here is the basic idea for a simple zero $\rho = \frac{1}{2} + i\gamma$. It is straightforward to modify for multiple zeros.
Write the Laurent expansions of each term around $\rho$, noting that for $\gamma \neq 0$ the only term with a zero is $\zeta(s)$:
\begin{align} \zeta(s) &= 0 + \zeta'(\rho)(s - \rho) + O\big( (s - \rho)^2 \big) \\ \zeta(2s) &= \zeta(2\rho) + 2\zeta'(2\rho)(s - \rho) + O\big( (s - \rho)^2 \big) \\ X^{s - \frac{1}{2}} &= X^{\rho - \frac{1}{2}} + (\log X) X^{\rho - \frac{1}{2}} (s - \rho) + O\big( (s - \rho)^2 \big) \\ s - \tfrac{1}{2} &= (\rho - \tfrac{1}{2}) + (s - \rho). \end{align}
The residue at $s = \rho$ will come from the coefficient of $(s - \rho)^{-1}$ in the Laurent expansion of
$$\frac{\zeta(2s)}{\zeta(s)} \frac{X^{s - \frac{1}{2}}}{s - \frac{1}{2}},$$
which direct insertion reveals comes from the constant terms in the Laurent expansions of $\zeta(2s), X^{s - \frac{1}{2}}$, and $(s - \tfrac{1}{2})$, and the coefficient of $\zeta(s)$. This shows that the residue is
$$ \frac{\zeta(2\rho)}{\zeta'(\rho)} \frac{X^{\rho - \frac{1}{2}}}{\rho - \frac{1}{2}} = \frac{\zeta(2\rho)}{\zeta'(\rho)} \frac{X^{i \gamma}}{i \gamma}. $$
For simple zeros, note that this is equivalent to taking the values at $s = \rho$ of the various nonzero, nonpolar terms, and multiplying by the residue of $\zeta(s)^{-1}$.