1
$\begingroup$

I know that $\zeta(2n)$ can be found using the contour integral $$\int_{\Gamma_N} \frac{cot(\pi z)}{z^{2n}}dz$$ and computing its residues as $N\to\infty$.

My question is could other values of the zeta function be derived through an integral such as $$\int_{\gamma_n} \frac{\Gamma(z)}{z^s}dz$$ since $\Gamma(z)$ has poles at every negative integer. I do have alot of doubts about it, since I am not sure that the contour will approach zero as $n\to\infty$, and I find it hard to believe that this was not already considered by someone else.

However, Wolfram Residue Calculator shows interesting values for the residues at zero for some s.

First few values for $s=1\to s=4$ are $$Res_{s=1}=-\gamma$$ $$Res_{s=2}=\frac{1}{12}(6\gamma^2+\pi^2)$$ $$Res_{s=3}=\frac{1}{12}(-2\gamma^3-\gamma\pi^2+2\psi^2(1))$$ $$Res_{s=4}=\frac{1}{480}(20\gamma^4+20\gamma^2\pi^2+3\pi^4-80\gamma\psi^2(1))$$ If these values do not have anything to do with $\zeta(s)$, can anyone at least provide some meaning to why these values appear and if there is any pattern/formula for deriving them?

$\endgroup$
  • $\begingroup$ The problem is that $|\Gamma(z)|$ does not vanish on the contour $\gamma_N$ as $N \to \infty$. $\endgroup$ – Ron Gordon Nov 1 '17 at 11:33
  • $\begingroup$ I see, but do these values have any connection to zeta function? Some of the values of look similar to different zeta values with the exception of the $\gamma$ and $\psi$. $\endgroup$ – aleden Nov 1 '17 at 11:50
  • $\begingroup$ Ignoring the divergence that makes this contour integral rather useless in that limit, you would end up with something more like a polylog rather than a zeta. $\endgroup$ – Ron Gordon Nov 1 '17 at 20:01
1
$\begingroup$

Well, it seems like this article and the reflection formula for the gamma function could explain your empirical observations. In particular, the reflection formula provides that $$\Gamma(x) = -\frac{\pi}{x \cdot \Gamma(-x) \sin(\pi x)},$$ where the Taylor series coefficients for the reciprocal gamma function (as in the article) satisfy the recurrence relation: $$n a_n = \gamma a_{n-1} - \zeta(2) a_{n-2} + \zeta(3) a_{n-3} - \cdots + (-1)^{n-1} \zeta(k), n > 2.$$ So this gives a rough estimation of how the Taylor series coefficients for $\Gamma(z)$ are related to the Riemann zeta function. Then the residues you mention above depend on these values. Note that $$\frac{1}{\sin(\pi x)} = \frac{1}{\pi x} + \frac{\pi x}{6} + \frac{7\pi^3 x^3}{360} + \frac{31 \pi^5 x^5}{15120} + \cdots. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.