Note: I'm refereshing my complex analysis skills in order to learn some analytic number theory. Here's one (basic) claim I'd like to prove and my attempt. My questions are:
- Is my partial attempt correct?
- Are there better (or shorter) ways to prove it? I may be going a bit too much into the details instead of using general theorems.
I'd like to prove that if $f:\mathbb{C}\rightarrow\mathbb{C}$ is defined by $f(z)=\int_1^\infty e^{-x}x^z\,dx$ then $f$ is complex analytic.
My attempt: Define $f_n(z)=\int_1^n e^{-x}x^z\,dx$. Then it's enough to prove that for each $n$, the function $f_n$ is holomoprphic and that $f_n$ converges uniformly on compact sets.
As for the first point, the integral is a limit of Riemann sums and each Riemann sum is complex analytic. So it's enough to prove that the Riemann sums converge to the integral uniformly in compact sets (relative to $z$). I think that follows from equicontinuity of $e^{-x} x^z$ in $[1,n]$ and $Re(z)$ being bounded in compact sets.
As for the second point, in a compact set $Re(z)$ is bounded, say, by $K$. Then $|x^z|\leq x^K$. Now $|f_m(z)-f_n(z)|= \left| \int_n^m e^{-x}x^z\,dx\right| \leq\int_n^m e^{-x}|x^z|\,dx \leq \int_n^m e^{-x}x^K \, dx \overset{n,m\rightarrow \infty}{\longrightarrow}0.$ So, the convergence really is uniform.