Show that there is no entire function $F(s)$ with $F(s)=O(e^{c|s|})$ that has simple zeros at $s=0,-1,-2,\cdots,$ and that vanishes no where else.

Question

An exercise from Stein's complex analysis related to Gamma Function

(a) Show that $1/|\Gamma(s)|$ is not $O(e^{c|s|})$ for any $c>0$;
(b) Show that there is no entire function $F(s)$ with $F(s)=O(e^{c|s|})$ that has simple zeros at $s=0,-1,-2,\cdots,$ and that vanishes no where else.

Following the hint(set $s=-k-1/2$) and using Stirling' s formula, I have worked out (a). For (b), my attempt is as follows. Suppose such $F$ do exists, since $\Gamma$ only has simple poles at $s=0,-1,-2,\cdots,$and vanishes no where else, than $F(s)\Gamma(s)$ is entire and vanishes nowhere, then $$ F(s)\Gamma(s)=e^{P(s)}. $$ If we can prove $P(s)=O(e^{d|s|})$ or give a similar estimate, then by (a),the problem is solved,but I stuck here.

Appreciate any help!

Answer 1

We can prove this by showing that any entire function of order one satisfying these characteristics cannot be $\mathcal O(e^{c|s|})$. Let $F(s)$ denote such function, so by Hadamard's factorization theorem we have

\begin{aligned} F(s) &=se^{A+Bs}\prod_{k=1}^\infty\left(1+\frac sk\right)e^{-s/k} \\ &=e^{A+(B-\gamma)s}\cdot se^{\gamma s}\prod_{k=1}^\infty\left(1+\frac sk\right)e^{-s/k} \\ &={e^{A+(B-\gamma)s}\over\Gamma(s)} \end{aligned}

Now, applying part (a) will give us the conclusion.