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Let us consider the function $$ x \mapsto \left| \frac{\Gamma(x+z+iy)}{\Gamma(x+iy)} \right|^2, $$ where $x,y,z \in \mathbb R$, $x \ge 1/2$, $z >0$, $i$ is the imaginary unit, and $\Gamma$ is the Euler gamma function.

Is this function non-decreasing?

My straightforward attempt to check this would be by computing the derivative, but this leads me to some impenetrable formulas.

Numerical evidence also suggests that the answer to the question is yes.

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    $\begingroup$ From the Weierstrass product, I got $$\left| {\frac{{\Gamma (x + z + iy)}}{{\Gamma (x + iy)}}} \right|^2 = \frac{{x^2 + y^2 }}{{(x + z)^2 + y^2 }}e^{ - 2\gamma z} \prod\limits_{k = 1}^\infty {\frac{{x^2 + 2xk + k^2 + y^2 }}{{(x + z)^2 + 2(x + z)k + k^2 + y^2 }}e^{2z/k} } . $$ Would this help? $\endgroup$
    – Gary
    Commented May 30, 2022 at 10:48
  • $\begingroup$ @Gary I don't know. Do you think this expression gives the monotonicity? $\endgroup$
    – user934318
    Commented May 30, 2022 at 13:23

3 Answers 3

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By the answers of Gary and River Li, it suffices to show that $$ U(x,z)=\sum_{k=0}^\infty \left(\frac{2(k+x)}{(k+x)^2+y^2} - \frac{2(k+x+z)}{(k+x+z)^2+y^2}\right)\ge 0 $$ for $x\ge\frac 12, z>0$, which, by Gary's computation of the difference, is equivalent to showing that $$ \sum_{k=0}^\infty \frac{(k+x)^2-y^2}{[(k+x)^2+y^2]^2}\ge 0 $$ for $x\ge\frac 12$ (this is just the derivative of the sum with respect to $z$ with $x+z$ replaced by $x$).

Denoting $w=x+iy$, we see that the last sum can be written as $$ \sum_{k=0}^\infty\Re\frac{1}{(w+k)^2}\,. $$ This series is dominated by $\sum_{k=0}^\infty\frac{1}{(k+\frac 12)^2}$ in the half-plane $\Re w\ge \frac 12$, so it defines a bounded harmonic function there. Thus it suffices to show that it is non-negative on the boundary $x=\Re w=\frac 12$.

However there it can be computed. First, note that $$ \sum_{k=0}^\infty \frac{(k+\frac 12)^2-y^2}{[(k+\frac 12)^2+y^2]^2} =\frac 12\sum_{k=-\infty}^\infty \frac{(k+\frac 12)^2-y^2}{[(k+\frac 12)^2+y^2]^2}\,. $$ Now consider the meromorphic function $$ g(w)=-\frac{\tan(\pi w)}{(w-iy)^2}\,. $$ It has poles at half-integers where the real parts of the residues are exactly what we want to sum, and tends to $0$ like $1/R^2$ on the circles $|w|=R$ with integer $R$, so the sum of all residues must be $0$. Thus, our sum is just minus the real part of the residue of $g$ at $iy$, i.e., $$ \Re\left[\frac d{dw}\tan(\pi w)\right]=\Re\left[\frac \pi {\cos^2(\pi iy)}\right]\,, $$ which is, indeed, some positive number ($\cos(\pi iy)$ is real and not $0$).

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  • $\begingroup$ You may even add that $\cos(\pi i y) =\cosh(\pi y)\geq 1$. $\endgroup$
    – Gary
    Commented Jun 6, 2022 at 10:51
  • $\begingroup$ Nice. Where is " by Gary's computation of the difference"? $\endgroup$
    – River Li
    Commented Jun 7, 2022 at 14:32
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    $\begingroup$ @RiverLi I refer to the expression with $2z$ in front of the fraction. If you divide by $z$ and let $z\to 0$, you'll get the derivative (to be honest what I called "the derivative" is actually "half the derivative" but that makes no difference, so I'll leave it as it is to avoid bumping the post to the top). $\endgroup$
    – fedja
    Commented Jun 7, 2022 at 15:51
  • $\begingroup$ Very nice! Could you further explain the following two points: (1) "so it defines a bounded harmonic function there"? (2) "where the real parts of the residues are exactly what we want to sum" $\endgroup$
    – Dal
    Commented Jun 8, 2022 at 7:32
  • $\begingroup$ @Dal (1) If you have a sequence $f_n$ of analytic functions in the domain $\Omega$ such that $|f_n|\le a_n$ in $\Omega$ and $\sum_n a_n<+\infty$, then the series $\sum_n f_n$ converges uniformly in $\Omega$ and the uniform limit of analytic functions is analytic. (2) is a straightforward computation (perhaps, I lost an irrelevant factor of $\pi$ somewhere). What do you get when you try it? $\endgroup$
    – fedja
    Commented Jun 8, 2022 at 11:18
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By Gary's comment, it is enough to show that for any $y,C,z\in\mathbb{R}^+$ the function

$$ f:x \mapsto \frac{(x+C)^2+y^2}{(x+C+z)^2+y^2} $$

is increasing on $\left[\frac{1}{2},+\infty\right)$. We may notice that

$$ \frac{f'(x)}{f(x)}=\frac{d}{dx}\left(\log f(x)\right) = \frac{2(C+x)}{(C+x)^2+y^2}-\frac{2(C+x+z)}{(C+x+z)^2+y^2}=2z\frac{(C+x)(C+x+z)-y^2}{((C+x)^2+y^2)((C+x+z)^2+y^2)}$$ so the monotonicity clearly holds if $x\geq |y|$, but it might fail otherwise.

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  • $\begingroup$ Thanks! Do you see any way of assuming only $y \in \mathbb R$ and $x \ge 1/2$ instead of the more restrictive $x \ge |y|$? $\endgroup$
    – user934318
    Commented May 31, 2022 at 11:23
  • $\begingroup$ @user934318: I do not see a straightforward way to prove the claim in the initial setting, but I guess that, if possible, to consider a logarithmic derivative is an important step. $\endgroup$ Commented May 31, 2022 at 11:37
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Some thoughts:

By Gary's comment, letting $$f(x) := \ln \left| {\frac{{\Gamma (x + z + iy)}}{{\Gamma (x + iy)}}} \right|^2,$$ we have $$f'(x) = \sum_{k=0}^\infty \left(\frac{2(k+x)}{(k+x)^2+y^2} - \frac{2(k+x+z)}{(k+x+z)^2+y^2}\right).$$

If $x^2 + xz \ge y^2$, we have $\frac{2(k+x)}{(k+x)^2+y^2} - \frac{2(k+x+z)}{(k+x+z)^2+y^2} \ge 0$ and $f'(x) \ge 0$.

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