Functional equation for Riemann zeta function

**Cairo** · March 29th 2011, 09:12 AM

Let

$f(s)=\frac{1}{2}s(s-1)\pi^{-\frac{s}{2}}\Gamma(\frac{1}{2}s)\zeta(s)$ .

Prove that $\underset{s\rightarrow o}{\lim}\, f(s)=\underset{s\rightarrow1}{\lim\,}f(s)=\frac{1} {2}$ , and that f has no zeros outside the strip $0\leq\sigma\leq1$ .

**chisigma** · March 29th 2011, 11:05 PM

Originally Posted by Cairo

Let

$f(s)=\frac{1}{2}s(s-1)\pi^{-\frac{s}{2}}\Gamma(\frac{1}{2}s)\zeta(s)$ .

Prove that $\underset{s\rightarrow o}{\lim}\, f(s)=\underset{s\rightarrow1}{\lim\,}f(s)=\frac{1} {2}$ , and that f has no zeros outside the strip $0\leq\sigma\leq1$ .

Remembering the 'Hadamard product'...

$\displaystyle \zeta(s)= \pi^{\frac{s}{2}}\ \frac{\prod_{p} (1-\frac{s}{p})}{2\ (s-1)\ \Gamma(1+\frac{s}{2})}$ (1)

... and taking into account that is...

$\displaystyle \Gamma(1+\frac{s}{2}) = \frac{s}{2}\ \Gamma(\frac{s}{2})$ (2)

... You obtain...

$\displaystyle f(s) = \frac{1}{2}\ \prod_{p} (1-\frac{s}{p})$ (3)

... where p are the 'non trivial zeroes' of $\zeta (*)$ and they all have real part $0 \le \sigma \le 1$ . From (3) is evident that is...

$\displaystyle \lim_{s \rightarrow 0} f(s)=\frac{1}{2}$ (4)

... and because is $f(s)=f(1-s)$ is also...

$\displaystyle \lim_{s \rightarrow 1} f(s)=\frac{1}{2}$ (5)

Kind regards

$\chi$ $\sigma$

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