gamma functionreference-requestspecial functions
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.
We have that $$ r_a(x) = {{\Gamma \left( {\left( {a + 1} \right)x} \right)} \over {\Gamma \left( x \right)}} = {{\Gamma \left( {x + a\,x} \right)} \over {\Gamma \left( x \right)}} = x^{\,\overline {\,a\,x\,} } $$
where $x^{\,\overline {\,y\,} } $ denotes the Rising Factorial.
Now, the Rising Factorial is defined (for $x$ and $y$ real and also complex) as $$ h(x,y) = x^{\,\overline {\,y\,\,} } = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}} = \prod\nolimits_{\;k\, = \,\,0}^{\,y} {\left( {x + k} \right)} $$ where the last term denotes the Indefinite Product, computed for $k$ ranging between the indicated bounds.
So we can write $f_a(x)$ as $$ \bbox[lightyellow] { f_{\,a} (x) = 2^{\,a\,x} {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}} = \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {2\left( {x + k} \right)} = 2^{\,a\,x} x^{\,a\,x} \prod\nolimits_{\;k\, = \,\,0}^{\,a\,x} {\left( {1 + k/x} \right)} } \tag{1}$$
Concerning the derivative of $r_a(x)$, since $$ \left\{ \matrix{ {\partial \over {\partial x}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\left( {\psi \left( {x + y} \right) - \psi \left( x \right)} \right) \hfill \cr {\partial \over {\partial y}}h(x,y) = {{\Gamma \left( {x + y} \right)} \over {\Gamma \left( x \right)}}\psi \left( {x + y} \right) \hfill \cr} \right. $$ then, as you already found $$ \eqalign{ & {d \over {dx}}r_a(x) = {\partial \over {\partial x}}h(x,y) + {\partial \over {\partial y}}h(x,y){d \over {dx}}y = \cr & = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {\left( {a + 1} \right)\psi \left( {x + ax} \right) - \psi \left( x \right)} \right) = \cr & = {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\left( {a\psi \left( {x + ax} \right) + \left( {\psi \left( {x + ax} \right) - \psi \left( x \right)} \right)} \right) \cr} $$
where, for $0<x$ (and $0<a$) $\Gamma(x+ax)/\Gamma(x)$ is clearly positive.
However, while $\psi(x+ax)-\psi(x)$ is also positive since $\psi(x)$ is increasing in that range, $a\psi(a+ax)$ introduces a negative term for lower $x$.
To determine the limit of $r_a'(x)$ as $x \to 0^+$, let's consider the series development of $$ \bbox[lightyellow] { \left\{ \matrix{ \ln \Gamma (cx) = \ln \left( {{1 \over {cx}}} \right) - \gamma cx + O\left( {x^{\,2} } \right) \hfill \cr \psi (cx) = - {1 \over {cx}} - \gamma + {{\pi ^{\,2} } \over 6}cx + O\left( {x^{\,2} } \right) = \hfill \cr = - {1 \over {cx}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + cx}}} \right)} \hfill \cr} \right. } \tag{2}$$
Therefore $$ \eqalign{ & \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {d \over {dx}}r_a(x) = \mathop {\lim }\limits_{x\; \to \;0^{\, + } } {{\Gamma \left( {x + ax} \right)} \over {\Gamma \left( x \right)}}\mathop {\lim }\limits_{x\; \to \;0^{\, + } } \left( {\left( {a + 1} \right)\psi \left( {x + ax} \right) - \psi \left( x \right)} \right) = \cr & = {1 \over {a + 1}}\left( { - \gamma a} \right) \cr} $$
which is negative for $0<a$.
Concerning $f_a(x)$ instead $$ \eqalign{ & {d \over {dx}}f_{\,a} (x) = \;2^{\,a\,x} a\ln 2r_{\,a} (x) + 2^{\,a\,x} {d \over {dx}}r_{\,a} (x) = \cr & = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \left( {r_{\,a} (x)} \right)} \right) = \cr & = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + {d \over {dx}}\ln \Gamma (x + ax) - {d \over {dx}}\ln \Gamma (x)} \right) = \cr & = 2^{\,a\,x} r_{\,a} (x)\left( {a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x)} \right) \cr} $$ (which is the equation you already found)
and $$ \bbox[lightyellow] { \eqalign{ & \mathop {\lim }\limits_{x\; \to \;0^{\, + } } f_{\,a} '(x) = 1\left( {\left( {a\ln 2} \right){1 \over {a + 1}} - {{\gamma a} \over {a + 1}}} \right) = \cr & = {a \over {a + 1}}\left( {\ln 2 - \gamma } \right) = {a \over {a + 1}}0.1159 \cdots \cr} } \tag{3}$$
Proceeding with the development of the derivative above $$ \bbox[lightyellow] { \eqalign{ & {d \over {dx}}f_{\,a} (x)\;\mathop /\limits_{} \;\left( {2^{\,a\,x} r_{\,a} (x)} \right) = \cr & = a\ln 2 + \left( {a + 1} \right)\psi (x + ax) - \psi (x) = \cr & = a\ln 2 + \left( {a + 1} \right)\left( { - {1 \over {\left( {a + 1} \right)x}} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)} } \right) - \left( { - {1 \over x} - \gamma + \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + x}}} \right)} } \right) = \cr & = \left( {a\ln 2 - {1 \over x} - \left( {a + 1} \right)\gamma + {1 \over x} + \gamma } \right) + \left( {a + 1} \right)\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + \left( {a + 1} \right)x}}} \right)} - \sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {1 \over {k + 1 + x}}} \right)} = \cr & = a\left( {\ln 2 - \gamma } \right) + \sum\limits_{0\, \le \,k} {\left( {{a \over {k + 1}} + {1 \over {k + 1 + x}} - {{\left( {a + 1} \right)} \over {k + 1 + \left( {a + 1} \right)x}}} \right)} = \cr & = a\left( {\ln 2 - \gamma } \right) + a\sum\limits_{0\, \le \,k} {\left( {{1 \over {k + 1}} - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}} \right)} \cr} } \tag{4}$$
and $$ \bbox[lightyellow] { \eqalign{ & 0 \le {1 \over {k + 1}} - {{k + 1} \over {\left( {k + 1 + x} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}} = \cr & = {1 \over {k + 1}} - {1 \over {\left( {1 + x/\left( {k + 1} \right)} \right)\left( {k + 1 + \left( {a + 1} \right)x} \right)}}\quad \left| {\;0 \le x,k} \right. \cr} } \tag{5}$$ therefore your thesis is demonstrated.
There's no known closed form for this. Any alternative expression still involves two gammas; say, using the duplication formula, you can rewrite it in terms of $\Gamma(x)$ and $\Gamma(x/2)$, but it doesn't really help. The "power law" holds only in the asymptotic sense, because of $\lim\limits_{x\to+\infty}\Gamma(x+a)/[x^a\Gamma(x)]=1$ for any real $a$ (in our case $a=1/2$).
Best Answer
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 $0$).