# $f(s)=\int_r ^\infty x^{s-1}/(e^x-1) dx$ is an entire function

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?

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