Show that $2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}$ is an increasing function

Question

I would like to show that the following function \begin{align} f_a(x)=2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)} \end{align} is an increasing function in $x$ for $x \ge 0$ for any fixed $a>0$.

I did some simulations but not sure how to show a proof for this. I wanted to point out that, from simulation, it seems that $\frac{\Gamma((a+1)x)}{\Gamma(x)}$ can be decreasing for values $x=0$.

We can attempt this by showing that the derivative of a logarithm of $f_a(x)$ is positive. Let

\begin{align} g_a(x)= \log (f_a(x)) \end{align}

(here log is base e) and the derivative of $g_a(x)$ is given by \begin{align} \frac{d}{dx} g_a(x)&= \frac{d}{dx} \left( ax \log(2)+ \log (\Gamma( (a+1)x))- \log (\Gamma( x)) \right)\\ &=a \log(2) + (a+1)\psi((a+1)x)-\psi(x) \end{align} where $\psi(x)$ is a digamma function.

Now it remains to show the following inequality for the difference of digamma functions \begin{align} (a+1)\psi((a+1)x)-\psi(x) \ge - a \log(2) . \end{align}

Answer 1

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.

Answer 2

At least for large $x$, considering $$f_a(x)=2^{ax}\frac{\Gamma((a+1)x)}{\Gamma(x)}$$ take logarithms $$\log(f_a(x) )=a x\log(2)+\log (\Gamma ((a+1) x))-\log (\Gamma ( x))$$ Now, use Stirling approximation for $\log(\Gamma(p))$ and continue with Taylor expansion for large $x$ and get $$\log(f_a(x) )=x \left(a \log \left(\frac{2x}{e}\right)+(a+1) \log (a+1)\right)+\frac{1}{2} \log \left(\frac{1}{a+1}\right)-\frac{a}{12 (a+1) x}+O\left(\frac{1}{x^4}\right)$$

Answer 3

Sketch of proof:

The ratio $G^a(x):=\Gamma(x+ax)/\Gamma(x)$ for $a>0$ is a function that has a single minimum for $x>0$, and the location of that minimum in the limit that $a \to 0$ (but not exactly zero) is the root of the equation $x\,\psi^{\,'}(x) + \psi(x)=0.$ Call this value $x_0$ which has the numerical value of approximately 0.216099. For $a>>0$, the minimum of $G^a(x)$ occurs for some $0<x^*<x_0.$ Since we want an inequality valid for $x \ge 0$ it is natural to look at $G_a(x)$ for $x \sim 0.$ A linear approximation is $G^a_1(x) = 1/(1+a) - a/(1+a)\gamma \, x,$ where $\gamma$ is Euler's constant. The linear approximation will undershoot the value $G^a(x^*)$ which is what will ultimately result in an inequality. To counteract the falling trend from 0 to $x^*$ we seek to find $y$ such that $$\frac{d}{dx} y^x\, G^a(x) |_{x=0} = 0 \text{ or its approx } \frac{d}{dx} y^x\, G^a_1(x) |_{x=0} = 0.$$ This is posed as a question involving a derivative because we want to find the best $y.$ Doing the last calculation it is found that $\log{y}=a\, \gamma.$ However $2^{a\,x}> (e^\gamma)^{a\,x}$ since $e^\gamma \sim 1.78$. Thus the proposed inequality is true and can be improved with 2 replaced by $e^\gamma$. To make this argument rigorous will take some work, but it may only need an appeal to Bohr-Mollerup.