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Going through a proof in Analyti number theory, the calculation of the residues of $$ f(s) = \frac{x^s}{s}\frac{\zeta'(s)}{\zeta\phantom{'}(s)} $$ came up. I do have some experience with complex analysis so I tried to compute the residues myself. So far it seems the residues are $$ \mathrm{Res}(0) = \log 2\pi \, , \ \mathrm{Res}(1) = -x \, , \ \mathrm{Res}(-2n) = -\frac{1}{n x^n} $$ where $n \in \mathbb{N}$. I have been able to calculate the resiudes at $0$ and $1$ using the asymmetric functional equation for $\zeta(s)$. But alternative approaches is still very welcome.

My biggest problem is dealing with the residues at $-2n$, do anyone have any suggestions?

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  • $\begingroup$ You may use the Hadamard product to obtain $$\frac{\zeta'(s)}{\zeta\phantom{'}(s)}=(\log\,\zeta(s))' =\cdots+\sum_{\rho}\left(\frac s{\rho}+\log\left(1-\frac s{\rho}\right)\right)'=\cdots+\sum_{\rho}\frac 1{\rho}+\frac 1{s-\rho}$$ Once multiplied by $\dfrac{x^s}s$ you'll get a residue $\dfrac{x^{\rho}}{\rho}\,$ at every zero (trivial or not). Hoping this clarified things, $\endgroup$ – Raymond Manzoni May 11 '14 at 15:10
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Observe that poles and zeroes of $\zeta$ are all simple poles for its logarithmic derivative, hence the singularities of $s\longmapsto\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}$ lie all in halfplane $\{\sigma\leq1\}$.

Now, the poles of this function are: the pole of $\frac{x^{s}}{s}$ ($s=0$) which residue is $$ Res\left(\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s},0\right) =\lim_{s\rightarrow1}\left(s\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}\right)=\frac{\zeta'(0)}{\zeta(0)}=\log2\pi\;, $$ and zeroes of $\zeta$ (trivial and non trivial) whose residue is \begin{align*} Res\left(\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s},\rho\right) =\lim_{s\rightarrow\rho}\left((s-\rho)\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}\right)=\frac{x^{\rho}}{\rho}\;, \end{align*} and the pole of $\zeta$ ($s=1$) which residue is $$ Res\left(\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s},1\right) =\lim_{s\rightarrow1}\left((s-1)\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}\right)=-x\;, $$ in fact, since $(s-1)\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s}=\frac{(s-1)^{2}\zeta'(s)}{(s-1)\zeta(s)}\frac{x^{s}}{s}$, being $s=1$ simple pole for $\zeta$, Laurent development around $1$ becomes $\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^{+\infty}c_{n}(s-1)^{n}$ which implies $\zeta'(s)=\frac{-1}{(s-1)^{2}}+\sum_{n=1}^{+\infty}nc_{n}(s-1)^{n-1}$. Then conclude by a simple computation.

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  • $\begingroup$ I am glad to see you agree with the residues I posted. But I sadly do not see any compelling - or any? - argument for the poles at $\rho$? Do you have a proof for the residues here? $\endgroup$ – N3buchadnezzar Apr 26 '14 at 13:44

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