5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $f(z) =\frac{\Gamma(z+a)}{\Gamma(z)}$ 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 $\frac{\Gamma(z + 1 + ix)\Gamma(z + 1 - ix)}{\Gamma(z + 1)\Gamma(z + 1)}$ $\endgroup$ – reuns May 10 '17 at 8:15
  • $\begingroup$ 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? $\endgroup$ – Sammyy Delbrin May 10 '17 at 8:19
  • $\begingroup$ 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 $\endgroup$ – reuns May 10 '17 at 8:28
  • $\begingroup$ Does that asymptotic expansion hold uniformly in $a$? Do you have a reference for me? $\endgroup$ – Sammyy Delbrin May 10 '17 at 8:32
  • $\begingroup$ It is not true as $a \to \infty$ but uniformly in $s$ (away from the poles of $\Gamma$) $\endgroup$ – reuns May 10 '17 at 8:36
6
+50
$\begingroup$

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.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.