# Gamma duplication formula via Hadamard

$$F(s) := \frac{\Gamma(s)\Gamma(s + 1/2)}{\Gamma(2s)} = \sqrt{\pi} 2^{1 - 2s}$$

This is an entire function of order 1.

Also, it does not vanish anywhere. Hence if we already knew that it was order $< 2$, by Hadamard's factorization theorem we could write

$$F(s) = e^{As + B}$$

and then evaluating $F$ at $0$ and $1/2$ we get the duplication formula. Is there any way to easily see that $F$ is order $<2$? It's straightforward for $\text{Re}(s) > 1/2$ but for the other side I keep on coming to some statement like "$\sin (\pi s) \Gamma(s)$ is bounded below by $1/n!$ in neighborhoods $U_n \ni -n$ of width uniform in $n$", which I'm not sure how to deal with.

• maybe by induction on $n \in \mathbb{Z}$ looking at $\frac{\Gamma(2(s+n))}{\Gamma(s+n) \Gamma(s+n+1/2)}$ – reuns Apr 4 '16 at 7:02
• Ah, ok, and prove that that function is of constant magnitude on vertical lines. – Julien Clancy Apr 4 '16 at 20:40
• I don't know, but I'm not 100% sure you need it, if you can prove that $F(s)$ or $1/F(s)$ (the two being shown entire) is bounded on some vertical strip $Re(s) \in [a,a+1[$ – reuns Apr 5 '16 at 1:33