complex analysis – Riemann zeta analytic continuation


$pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}, Re(s)>1, where theta(u)=sum_{n=-infty}^{infty}e^{-n^2pi u}$.

$pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$= $int_0^1 frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$+$int_1^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$.
Transforming u to 1/u in the first integral and using $theta(1/u)=u^{1/2}theta(u)$
We get,
$pi^{-s/2}Gamma(s/2)zeta(s)=int_1^infty frac{(u^{1/2}theta(u)-1)}{2}u^{(s-1)/2}frac{du}{u}$+$int_1^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$
So the R.H.S. is analytic for all s such that $0<Re(s)<1$. So is the first integral $pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$ valid for $0<Re(s)<1$?