Thread: Functional equation for Riemann zeta function

Let

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

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

Originally Posted by Cairo

Let

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

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

Remembering the 'Hadamard product'...

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

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

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

... You obtain...

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$