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For the Riemann zeta function, we know of the standard functional equation that relates $\zeta(s)$ and $\zeta(1-s)$. I wanted to know whether there are functional equations that relates $\zeta(s)$ and $\zeta(s-1)$?

EDIT: My main motivation behind asking this question is I have found such an equation, but I do not know whether such an equation exists in literature. Also, I do not want to appear as if I am promoting my formula here, but rather I am more interested in the works that have been done in such directions.

As per @lhf's request here is my formula, for $\Re(s) > 1$ $$ \zeta(s) + \frac{2}{s-1}\zeta(s-1) = \frac{s}{s-2} - s\int_1^\infty \frac{\{x\}^2}{x^{s+1}} dx$$ where $\{x\}$ is the fractional part of x.

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In the conference volume Algebraic Number Fields: L-functions and Galois Properties (1977), there is an article by Audrey Terras with the title "A relation between $\zeta_K(s)$ and $\zeta_K(s-1)$ for any algebraic number field $K$". –  KCd May 20 '12 at 16:02
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Just show us your equation and ask whether it is known. It is probably a simpler question to answer than the current one. –  lhf May 20 '12 at 16:03
    
@lhf "I am more interested in the works that have been done in such directions" (OK I will post it, not a big deal, except the part that it might be incorrect. So, please let me check that before I post it!). –  Roupam Ghosh May 20 '12 at 16:39
    
@KCd Thanks I will check that out. –  Roupam Ghosh May 20 '12 at 16:39

3 Answers 3

Yes, $\zeta(\overline{s})=\overline{\zeta(s)}$ for $s \in \mathbb C-\{1\}$.

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Unfortunately, we don't have a functional equation connecting $\zeta(s)$ and $\zeta(s-1)$. The reason being that $\{x\}$ is not perfectly periodic and cannot be developed into a Fourier series.

In case you are familiar with the Euler-Maclaurin summation, I can give a brief idea of how we have \begin{equation} \zeta(s)=2(2\pi)^{s-1}\Gamma(1-s)\sin\left(\frac{\pi s}{2}\right)\zeta(1-s). \end{equation}

From Euler-Maclaurin, we have

\begin{align} \zeta(s) & = \frac{1}{s-1}+\frac{1}{2}+\sum_{r=2}^{q}\frac{B_r}{r!}(s)(s+1)\cdots(s+r-2) \\ & \phantom{=} -\frac{(s)(s+1)\cdots(s+q-1)}{q!}\int_{1}^{\infty}B_{q}(x-[x])x^{-s-q} ~dx \end{align}

here $B_{q}(x-[x])$ are the periodic Bernoulli polynomials and have a Fourier expansion for $q\geq 2$.

Putting $q=3$, and using Fourier expansion, we can have the Riemann functional equation.

Just to make a slight correction, your formula is valid for $\Re(s)>2$ as

\begin{equation} -s\int_{1}^{n}\frac{(t-[t])^{2}}{t^{s+1}}dt=\sum_{k=1}^{n-1}\int_{k}^{k+1}(t-k)^{2}d(t^{-s})=\\\sum_{k=1}^{n-1}[(k+1)^{-s}+\frac{2}{s-1}(k+1)^{1-s}]+\frac{2}{(s-1)(s-2)}(n^{2-s}-1) \end{equation}

But since the integral

\begin{equation} -s\int_{1}^{\infty}\frac{(t-[t])^{2}}{t^{s+1}}dt \end{equation}

converges absolutely in the plane $\Re(s) >0$, therefore you can have your formula valid for $\Re(s)>0$ {analytic continuation} (excluding the poles $s=2$ and $s=1$ in the equation).

PS: I think you should try again for Indian Statistical Institute, Chennai Mathematical Institute and Institute of Mathematics and Applications next year. Where are you studying currently?

Best Wishes,

Sumit

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Hey Sumit, I haven't really looked a long time here at Math.SE, due to work. You can find my bigger post on this topic here at mathoverflow.net/questions/97929/… And thanks for the encouragement. Right now I am really engrossed with my day job, cause I am not studying. But IMO maths research is a very personal thing, and whether or not you go into a premier institute, you can still explore the mysteries still beyond our understanding. I would be happy to meet you sometime. Cya! –  Roupam Ghosh Nov 6 '12 at 8:10

Does it have to be a relation between $\,\zeta(s)\,,\,\zeta(s-1)\,$? If it doesn't then you can have, say $$\,\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}\,\,,\,\,with\,\,\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$$ which extends the zeta function to $\,\operatorname{Re}(s)>0\,\,,\,s\neq 1$

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Yes, I was thinking in the direction of $\zeta(s)$ and $\zeta(s-1)$. Maybe I did not express myself clearly, because I know these two answers that have been posted. I will edit my question a bit. Thanks for your answers though. :) –  Roupam Ghosh May 20 '12 at 15:41

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