# Talk:Riemann zeta function

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## Gamma Function definition

In the Riemann zeta function page it is written that:

${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {1}{e^{x}-1}}\,x^{s}{\frac {\mathrm {d} x}{x}}}$

and that:

${\displaystyle \Gamma (s)=\int _{0}^{\infty }e^{-x}\,x^{s}{\frac {\mathrm {d} x}{x}}}$.

However looking at the Gamma function definition in Wikipedia one can see that:

${\displaystyle \Gamma (s)=\int _{0}^{\infty }e^{-x}\,x^{s-1}{\frac {\mathrm {d} x}{x}}}$.

Correspondingly ${\displaystyle \zeta (s)={\frac {1}{\Gamma (s)}}\int _{0}^{\infty }{\frac {1}{e^{x}-1}}\,x^{s-1}{\frac {\mathrm {d} x}{x}}}$.

More looking at Riemann's paper he has the s-1 also.

However Riemann instead of Gamma used factorial ${\displaystyle \Gamma (n)=(n-1)!=\Pi (n-1)}$.

I tried to fix the page but it was rejected as vandalism.

Please correct it or explain to me what am I wrong about.

Adikatz (talk) 06:52, 26 July 2017 (UTC)

In the definition of the zeta function the definition of the Gamma function is wrong. It should be s-1 instead of s. The s-1 should also be in the integral representing the gamma function. I have tried to fix but it was revoked as vandalism.

Vandalism is "malicious or ignorant destruction" (Webster's). In your case the second definition occurs. The gamma function is defined as

${\displaystyle \Gamma (s)=\int _{0}^{\infty }e^{-x}\,x^{s-1}{\mathrm {d} x}}$. Or ${\displaystyle \Gamma (s)=\int _{0}^{\infty }e^{-x}\,x^{s}{\frac {\mathrm {d} x}{x}}}$. But not ${\displaystyle \Gamma (s)=\int _{0}^{\infty }e^{-x}\,x^{s-1}{\frac {\mathrm {d} x}{x}}}$.Sapphorain (talk) 07:57, 26 July 2017 (UTC)

Sorry. I was mistaken since I did not pay attention to the fact that the dx was divided by x. Adikatz (talk) 14:21, 26 July 2017 (UTC)

## Simple Approximation

${\displaystyle \zeta (x)\approx {\begin{cases}{\frac {1}{x-1}}+\gamma &{\mbox{if }}1\leq x\leq 2\\\\\pi ^{x}/(3{\sqrt {10}}^{x-1})&{\mbox{if }}x>2\end{cases}}}$

Also, for ${\displaystyle x>1}$ we have:

${\displaystyle \lim _{x\to 1}[\zeta (x)-{\frac {1}{x-1}}]=\gamma }$

where, in both expressions, ${\displaystyle \gamma }$ refers to the Euler-Mascheroni constant. — Craciun Lucian.

## Anon edit needs vetting

The following anonymous edit comes from an IP with a very checkered history. It needs vetting:

${\displaystyle -\zeta (s)=\prod _{p}{\frac {1}{1-p^{-s}}}+\zeta (s)=\prod _{p\in primes}{\frac {1}{1-p^{-s}}}}$

Thanks. --Wetman 13:02, 18 Apr 2005 (UTC)

This is surely wrong, as it implies that the \zeta function is zero. Oleg Alexandrov 18:15, 18 Apr 2005 (UTC)
I don't see this exact thing in the article though. Oleg Alexandrov 18:18, 18 Apr 2005 (UTC)
Wetman posted a diff; he is talking about the edit from 9:32, 8 April 2005, which Oleg reverted a few hours later anyway. No matter, the edit was fine, the product is over primes. Case closed. linas 22:09, 18 Apr 2005 (UTC)

## Perfect powers

Is there a formula out there that relates the zeta function to the perfect powers (1, 4, 8, 9, 16, 25, 27, 32, etc) in a similar way that Euler's product formula does for primes? The reason I ask is because I discovered a very simple one a while back, but I can't find information about any other such formulas. Thanks. --Vagodin 14:47, August 21, 2005 (UTC)

## Globally convergent series

The 'globally convergent series' found by Hasse appears to be essentially just the Euler transform applied to the Dirichlet eta function.

I programmed the second of the two Hasse series, and the second one seems to be wrong. It returns large incorrect values. If there's a bug, I must be blind:

The first series (which works correct) uses the following code:

double hasse0(double s, uint64_t limit) {

 double u = 0;
for (uint64_t n=0; n<limit; ++n) {
double r = 0;
for (uint64_t k=0; k<=n; ++k) {
double t = choose(n,k) / pow(k+1,s);
if (k%2) r -= t; else r += t;
}
u += r/pow(2,n+1);
}
return u/(1-pow(2,1-s));


}

The second series (which does not work) uses the exact same template:

double hasse1(double s, uint64_t limit) {

 double u = 0;
for (uint64_t n=0; n<limit; ++n) {
double r = 0;
for (uint64_t k=0; k<=n; ++k) {
double t = choose(n,k) / pow(k+1,s-1);
if (k%2) r -= t; else r += t;
}
u += r/(n+1);
}
return u/(s-1);


}

## Some more representations

Here are some more representations, but I don't know what subsection to put them in, or what commentary to give on them:

${\displaystyle \zeta (n)=\underbrace {\int _{0}^{1}\int _{0}^{1}\ldots \int _{0}^{1}} _{n}\,{\frac {\mathrm {d} x_{1}\mathrm {d} x_{2}\ldots \mathrm {d} x_{n}}{1-x_{1}x_{2}\ldots x_{n}}}}$ for positive integer n

${\displaystyle -{\frac {\zeta (s)}{s}}=\int _{0}^{\infty }\operatorname {frac} \left({\frac {1}{t}}\right)t^{s-1}\mathrm {d} t}$ for 0 < Re(s) < 1, where frac is the fractional part

${\displaystyle \Gamma (s+2)\zeta (s+2)=\int _{0}^{1}\int _{0}^{1}{\frac {[-\ln(xy)]^{s}}{1-xy}}\,\mathrm {d} x\,\mathrm {d} y}$ for Re(s) > 1

--AndreRD (talk) 16:29, 27 July 2017 (UTC)

## Graph

Under "Specific values", the graph seems to be of three functions, only one of which is the Zeta function. The other two seem to be based on a finite number of terms of the infinite series in the definition of the Zeta function.

## If s=1/2 in Riemann functional equation then the calculated result seems to disprove Riemann's hypothesis

${\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\ \sin \left({\frac {\pi s}{2}}\right)\ \Gamma (1-s)\ \zeta (1-s),}$

If the complex number s = 1/2 + 0j is put into the Riemann functional equation after the equation is rearranged so the left hand side reads E{s}/E{1-s} then the left hand side with a value of 1 does not equal the right hand side and this means that the complex part is not zero and so there can be no solution on the 1/2,0 coordinate of the complex plane as Riemann says there is. Have I got this right?

Soopdish (talk) 12:48, 8 August 2017 (UTC)Soopdish

The conjecture does not say "every point on the critical line is a zero", it says "every zero is on the critical line". So the fact that the point you looked at is nonzero is uninteresting and irrelevant. —David Eppstein (talk) 16:29, 8 August 2017 (UTC)