Riemann zeta function is one of the most mysterious functions that we encounter in mathematics.We require a meromorphic continuation of this function to ${\rm Re}(s)>0$ in order to prove the prime number theorem.Using partial summation formula we can deduce that for ${\rm Re}(s)>1$ we have $\zeta(s)=\frac{s}{s-1}-s\int_1^\infty x^{-s-1}\{x\}dx$ and we note that the right hand side makes sense for ${\rm Re}(s)>0$ and $s\neq 1$.
If we could show that the right hand side is meromorphic on ${\rm Re}(s)>0$ then we obtain a meromorphic continuation of $\zeta(s)$ to the half plane ${\rm Re}(s)>0$ and it is unique by the uniqueness theorem in complex analysis. In the notes given by our analytic number theory instructor, it is written that $F(s)=s\int_1^\infty x^{-s-1}\{x\}dx$ is meromorphic in the half plane ${\rm Re}(s)>0$ since the integral $\int_1^\infty |x^{-s-1}\{x\}|dx$ is uniformly convergent on every compact subset of the the half plane ${\rm Re}(s)>0$. Hence the right hand side represents a meromorphic function in the half plane ${\rm Re}(s)>0$ with a pole at $s=1$.
Having a 'not so strong' hold on complex analysis, I am having problem in understanding which theorem exactly I am using in this step and uniform convergence of the integral is with respect to which variable? Can someone provide me the exact statement we are using and the meaning of uniform convergence of integrals?
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You wrote
While that is how things happen in practice, logically speaking one does not need $\zeta(s)$ all the way out to ${\rm Re}(s) > 0$ to prove the Prime Number Theorem. All that's needed about $\zeta(s)$ is its analytic continuation to ${\rm Re}(s) \geq 1$ except for a simple pole at $s = 1$, plus $\zeta(s) \not= 0$ when ${\rm Re}(s) = 1$.
You also wrote that $F(s) = s\int_1^\infty (\{x\}/x^{s-1})\,dx$ "is meromorphic in the half plane ${\rm Re}(s) > 0$". In fact it is holomorphic there. The pole for $\zeta(s)$ at $s = 1$ comes from the term $1/(s-1)$ added to that integral.
Here is a way to prove $F(s)$ is analytic on ${\rm Re}(s) > 0$. Let $F_n(s) = \int_n^{n+1} (\{x\}/x^{s+1})\,dx$, so \begin{equation}\label{zeta1s} \int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx = \sum_{n \geq 1} F_n(s). \end{equation} For ${\rm Re}(s) > 1$, $F_n(s)$ is \begin{eqnarray*} \int_n^{n+1} \frac{\{x\}}{x^{s+1}}\,dx & = & \int_n^{n+1} \frac{x - n}{x^{s+1}}\,dx \\ & = & \frac{1}{s-1}\left(\frac{1}{n^{s-1}} - \frac{1}{(n+1)^{s-1}}\right) - \frac{n}{s}\left(\frac{1}{n^s} - \frac{1}{(n+1)^s}\right), \end{eqnarray*} which is an entire function (including at $s = 0$ and $s=1$). For all $s \in \mathbf C$,
$$ |F_n(s)| \leq \int_n^{n+1} \frac{dx}{x^{\sigma+1}} \leq \int_n^{n+1} \frac{dx}{n^{\sigma+1}} = \frac{1}{n^{\sigma+1}}, $$ where $\sigma = {\rm Re}(s)$, and $\sum F_n(s)$ converges (absolutely and) uniformly on ${\rm Re}(s) \geq \varepsilon > 0$, so on compact subsets of ${\rm Re}(s) > 0$. So on ${\rm Re}(s) > 0$, $\sum F_n(s)$ and $\zeta(s) - 1/(s-1)$ are analytic because a uniform limit of analytic functions on compact subsets is analytic.