# $f(z)=\int_1^\infty e^{-x}x^zdx$ is complex analytic

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:

1. Is my partial attempt correct?
2. 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 $f(z)=\int_1^\infty e^{-x}x^zdx$ then $f$ is complex analytic.

My attempt: Define $f_n(z)=\int_1^n e^{-x}x^zdx$. 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 $|f_m(z)-f_n(z)|=|\int_n^m e^{-x}x^zdx|\leq\int_n^m e^{-x}|x^z|dx \leq \int_n^m e^{-x}x^Kdx \overset{n,m\rightarrow \infty}{\longrightarrow}0$. So, the convergence really is uniform.

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I think on the one hand my answer is simpler than David Giraudo's, although on the other hand, it relies on some theorems, the inclusion of whose proofs would make it more complicated. – Michael Hardy Aug 2 '12 at 18:32

To see that $f_n$ is holomorphic, write $$x^z=\exp(z\log x)=\sum_{j=0}^{+\infty}z^j\frac{(\log x)^j}{j!}.$$ This series is normally convergent on $[1,n]$, hence we can switch the limit and the integral: $$f_n(z)=\sum_{j=0}^{+\infty}\left(\int_1^ne^{—x}(\log x)^jdx\right)\frac{z^j}{j!}.$$ Since $f_n$ is expressed as a power series, it's holomorphic.

What you have done shows that $\{f_n\}$ is uniformly convergent on compact sets, but not that the limit is $f$. To see that, write where $\Re z$ is bounded by $M$ that \begin{align} |f(z)-f_n(z)|&=\left|\int_n^{+\infty}e^{-x}\exp((\Re z+i\Im z)\log x)dx\right|\\ &\leq \int_n^{+\infty}e^{-x}\exp(\Re z\log x)dx\\ &\leq \int_n^{+\infty}e^{—x}e^{M\log x}dx\\ &=\int_n^{+\infty}e^{-x}x^Mdx, \end{align} and conclude using the convergence of $\int_1^{+\infty}e^{—x}x^Mdx$.

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I'd use Morera's theorem.

Morera's theorem says if $$\int_\gamma f(z)\,dz=0\tag{1}$$ for every simple closed curve $\gamma$, then $f$ is holomorphic.

Since the function is defined by an integral, the integral in $(1)$ becomes an iterated integral. You can then show that the hypotheses of Fubini's theorem are satisfied, so you can change the order of integration. Then conclude that what has now become the inside integral evaluates to $0$ because then you're integrating a function that you already know is holomorphic, along a closed curve.

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