# Is this product of Gamma functions bounded?

Consider the following term:

$$q(z,x):=\bigg\vert\frac{\Gamma(\sqrt{z} + 1 + ix)\Gamma(\sqrt{z} + 1 - ix)}{\Gamma(\sqrt{z} + 1)\Gamma(\sqrt{z} + 1)}\bigg\vert\cdot e^{\pi x}.$$

I would like to know whether the function $q$ is bounded for all values $x\in\mathbb{R}$ and $z=a+ib$ with $a>0$ fixed and $b\in\mathbb{R}$. That is, I want to know if the following statement is correct:

$$\exists\, c,a>0\,\forall x,b\in\mathbb{R}: \vert q(z,x) \vert \leq c,\, \text{where }z=a+ib.$$

Unfortunately, I could not find any reference for such an estimate and it indeed seems to be a tough problem. How could one analyze that function? In case the above statement is not true: What would be the best estimate one can hope for?

I would very appreciate any help.

• $f(z) =\frac{\Gamma(z+a)}{\Gamma(z)}$ is bounded away from its poles (at $-(z+a) \in \mathbb{N}$). For $x$ look at $\pi=\Gamma(s)\Gamma(1-s)\sin(\pi s)$. You can also look at the logarithmic derivative of $\frac{\Gamma(z + 1 + ix)\Gamma(z + 1 - ix)}{\Gamma(z + 1)\Gamma(z + 1)}$ – reuns May 10 '17 at 8:15
• Thank you. Can you explain what you mean by "For x look at"? Further, I can't see how to bring my term into the form $\Gamma(s)\Gamma(1-s)$ for some $s$. Can you give me a hint? – Sammyy Delbrin May 10 '17 at 8:19
• If $a > 0$ then $\Gamma(s+a) \sim s^a \Gamma(s)$, you can use the logarithmic derivative $\frac{\Gamma'(s)}{\Gamma(s)} = -\gamma-\sum_{n=0}^\infty (\frac{1}{s+n}-\frac{1}{n+1})$ en.wikipedia.org/wiki/Digamma_function – reuns May 10 '17 at 8:28
• Does that asymptotic expansion hold uniformly in $a$? Do you have a reference for me? – Sammyy Delbrin May 10 '17 at 8:32
• It is not true as $a \to \infty$ but uniformly in $s$ (away from the poles of $\Gamma$) – reuns May 10 '17 at 8:36

Complex transformations

Let $$\sqrt z = u + iv,$$ then $$a = u^2-v^2,\quad b = 2uv,\quad u\ge 0,\tag1$$ $$q(u,v,x) = \left|\Gamma(u+1+i(v+x))\Gamma(u+1+i(v-x))\over\Gamma^2(u+1+iv)\right|e^{\pi x}.$$ Let us use the formula $$\left|\Gamma(x+iy)\over\Gamma(x)\right|^2 = \prod_{n=0}^{\infty}\left(1+{y^2\over (x+n)^2}\right)^{-1}$$ (M. Abramowitz and I. A. Stegun. Handbook of mathematical functions), then $$\left|\Gamma(u+1+i(v\pm x))\over\Gamma(u+1)\right|^2 = \prod_{n=1}^{\infty}\left(1+{(v\pm x)^2\over (u+n)^2}\right)^{-1},$$ $$\left|\Gamma(u+1+iv))\over\Gamma(u+1)\right|^2 = \prod_{n=1}^{\infty}\left(1+{v^2\over (u+n)^2}\right)^{-1},$$ then $$q^2(u,v,x) = e^{2\pi x}\prod_{n=1}^{\infty}{(u+n)^2 + v^2\over(u+n)^2 + (v+x)^2}\cdot{(u+n)^2 + v^2\over(u+n)^2 + (v-x)^2}$$ $$= e^{2\pi x}\prod_{n=1}^{\infty}{\left((u+n)^2 + v^2\right)^2\over (u+n)^4 + 2(v^2+x^2)(u+n)^2 + (x^2-v^2)^2}$$ $$\ge e^{2\pi x}\prod_{n=1}^{\infty}{\left((u+n)^2 + v^2\right)^2\over (u+n)^4 + 2|v^2-x^2|(u+n)^2 + (v^2-x^2)^2}$$ $$\ge e^{2\pi x}\prod_{n=1}^{\infty}{\left((u+n)^2 + v^2\right)^2\over (u+n)^2 + |v^2-x^2|)^2}.$$ Easy to see that $q(u,v,0)=1.$

At the same time, $$q(u,v,x) \ge e^{\pi x}\quad \text{when}\quad x^2 < 2v^2.$$

So for the arbitrary case the answer is: NO.

For real $z$

Let us consider the situation $$z\in\mathcal R,\quad q = \sqrt z.$$ Then $$q(u,x) = \left|\Gamma(u+1+ix))\Gamma(u+1-ix))\over\Gamma^2(u+1)\right|e^{\pi x} = e^{\pi x}\prod_{n=0}^{\infty}\left(1+{x^2\over (u+1+n)^2}\right)^{-1},$$ $$\log q(u,x) \le \pi x - \sum_{n=0}^{\infty} \log\left(1+{x^2\over (u+1+n)^2}\right) \le \pi x - \sum_{n=1}^{\infty} {x^2\over (u+n)^2}\le \pi x - x^2\sum_{n=1}^{\infty} {1\over n^2} = \pi x - {\pi^2\over6}x^2 \le \pi{3\over\pi} - {\pi^2\over6}{9\over\pi^2} = {3\over2}.$$

So for real $z$ the answer is: YES.