16
$\begingroup$

Referring to this question I asked on math.SE. I am posting a more generalized question here, for answers and further inquiry.

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 $\zeta(s+1)$, or $\zeta(s)$, $\zeta(s+1)$, and $\zeta(s+2)$ or in general $\zeta(s)$, $\zeta(s+1)$, $\zeta(s+2)$, ..., $\zeta(s+n)$ for $\Re(s) > 0$ and for $n \in \mathbb{N}$?

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.

My formula goes as follows: For any $n \in \mathbb{N}$ and $\Re(s) > 0$ we have $$\frac{1}{s-1} + \sum_{r=1}^n \binom{n}{r} (-1)^r \left(\frac{\zeta(s+r-1)}{s+n-1} + \int_1^\infty\frac{\rho(x)^r}{x^{s+r}}\mathrm{d}x\right) = 0$$ where $\binom{n}{r}$ is the binomial coefficient and $\rho(x)$ is the fractional part of $x$

For example, putting $n=1$ we get the well known identity, $$\frac{1}{s-1} - \frac{\zeta(s)}{s} - \int_1^\infty\frac{\rho(x)}{x^{s+1}}\mathrm{d}x = 0$$ putting $n=2$, $$\frac{1}{s-1} - 2\left(\frac{\zeta(s)}{s+1} + \int_1^\infty\frac{\rho(x)}{x^{s+1}}\mathrm{d}x\right) + \left(\frac{\zeta(s+1)}{s+1} + \int_1^\infty\frac{\rho(x)^2}{x^{s+2}}\mathrm{d}x\right) = 0$$ and so on...

EDIT: Classic answer by Juan. This question is now solved.

$\endgroup$
0

1 Answer 1

16
$\begingroup$

Equations of this type are known. You may see, for example, the classical book "Primzahlen" by Landau paragraph 67. "Continuation of the zeta function by partial integration"

There it is proved the formula $$ (s-1)(\zeta(s)-1)-1=-\frac{(s-1)s}{2!}(\zeta(s+1)-1)- \frac{(s-1)s(s+1)}{3!}(\zeta(s+2)-1)-\cdots$$ $$\cdots-\frac{(s-1)s\cdots(s+q)}{(q+2)!}(\zeta(s+q+1)-1) -\frac{(s-1)s\cdots(s+q+1)}{(q+2)!}\int_1^\infty \frac{\rho(x)^{q+2}}{x^{s+q+2}} dx.$$

From this we may get the beautiful formula $$1=\binom{s-1}{1}(\zeta(s)-1)+\binom{s}{2}(\zeta(s+1)-1)+\binom{s+1}{3}(\zeta(s+2)-1)+ \cdots$$ (See Titchmarsh 2.14, formula (2.14.1) and also (2.14.2))

From the formula in Landau you see that the integral $\int_1^\infty \frac{\rho(x)^r}{x^{s+r}}\,dx$ can be expressed in terms of the $\zeta(s+k)$.

$\endgroup$
4
  • $\begingroup$ I must say that you integrates $\rho(x)^r$, I have not notice this before. Therefore, now I do not see any way to get your result from that of Landau. But, these are certainly functional equations of the type you wanted. $\endgroup$
    – juan
    Commented May 28, 2012 at 8:37
  • $\begingroup$ Yes, I have noticed that too, there are no $\rho(x)^r$ terms. Thanks for your answer anyways! $\endgroup$ Commented May 28, 2012 at 10:29
  • 1
    $\begingroup$ I have the book, but I followed your link. Then I realized that my formula was not well copied. In fact the formula of Landau has a power of $\rho(x)$ on the integrand. I have corrected it. Now I think your formula can be obtained from that of Landau. $\endgroup$
    – juan
    Commented May 28, 2012 at 14:54
  • $\begingroup$ Here's a direct link to the book by Landau mentioned by Juan. This takes you to paragraph 67. archive.org/stream/handbuchderlehre01landuoft#page/270/mode/2up $\endgroup$ Commented May 28, 2012 at 14:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .