The Gamma duplication formula reads

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

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.

  • $\begingroup$ maybe by induction on $n \in \mathbb{Z}$ looking at $\frac{\Gamma(2(s+n))}{\Gamma(s+n) \Gamma(s+n+1/2)}$ $\endgroup$ – reuns Apr 4 '16 at 7:02
  • $\begingroup$ Ah, ok, and prove that that function is of constant magnitude on vertical lines. $\endgroup$ – Julien Clancy Apr 4 '16 at 20:40
  • $\begingroup$ 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 $Re(s) \in [a,a+1[$ $\endgroup$ – reuns Apr 5 '16 at 1:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.