Let $r$ be a fixed positive real number, and for $s \in \Bbb C$, let $f(s)=\int_r ^\infty x^{s-1}/(e^x-1) dx$. Why $f$ is an entire function?

I tried to think as follows:

To use Morera's theorem, let $C$ be any closed curve in $\Bbb C$. We have to show that $\int_C f(s)ds=0$. But, I cannot change the order of two integrals, since one is an improper integral. How can I handle this?

  • $\begingroup$ I'm really confused. $z$ appears nowhere in your integral. $\endgroup$ – Cameron Williams Nov 20 '19 at 13:25
  • $\begingroup$ @CameronWilliams Sorry I'll make an edit $\endgroup$ – Quadr Nov 20 '19 at 13:27
  • $\begingroup$ Ah yep that's what I figured you meant, but wanted to be sure! Thanks for the edit. $\endgroup$ – Cameron Williams Nov 20 '19 at 13:28

Hint: let $f_n(s)=\int_r^{n} \frac {x^{s-1}} {e^{x}-1}dx$. Use Morera's Theorem to show that $f_n$ is entire. If we show that $f_n \to f$ uniformly on compact sets it would follows that $f$ is entire. Now $|f_n(s)-f(s)| \leq \int_n^{\infty} \frac {x^{M}} {e^{x}-1}dx$ for some constant $M$ as along as $s$ remains bounded (and $r>1$). I will let you prove the fact that $\int_n^{\infty} \frac {x^{M}} {e^{x}-1}dx \to 0$ as $n \to \infty$


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