Let $0<a \le 1, \alpha<0$ and $\beta>0$. How to prove that the function: $$f(x)=\frac{(\Gamma(a)-\Gamma(a,\alpha \ln(\beta x))) (\alpha\ln(x))^a}{(\alpha\ln(\beta x))^a (\Gamma(a)-\Gamma(a,\alpha \ln(x)))},$$ is decreasing for $\beta <1$ and increasing for $\beta>1$.
This question is motivated by the following inequality after drawing the graph for some values with wolfram. I tried the sign of derivative but it is more delicate.
Spoiler: With the approach suggested in the answer in the link, the result can only be proven for $\beta<1$. At least, this method does not allow to prove the result for $\beta>1$.
Let $g:\mathbb{R}\to\mathbb{\mathbb{C}}$ defined by $$ g(x)=\ln\left(\frac{\Gamma(a)-\Gamma(a,x)}{x^{a}}\right) $$
and let $h:\mathbb{R}\to\mathbb{C}$ defined by $$ h(x)=\Gamma(a)-\Gamma(a,x) $$
When $x\ge0$, both $g(x),h(x)\in\mathbb{R}$. When $x<0$, we can prove $g(x)\in\mathbb{R}$: \begin{eqnarray*} \frac{\Gamma(a)-\Gamma(a,x)}{x^{a}} & = & \frac{\int_{0}^{x}s^{a-1}e^{-s}ds}{x^{a}}\\ & = & \frac{\int_{0}^{x}|s|^{a-1}(e^{-i\pi})^{a-1}e^{-s}ds}{|x|^{a}(e^{-i\pi})^{a}}\\ & = & \int_{0}^{x}\left|\frac{s}{x}\right|^{a}\frac{e^{-ia\pi}}{e^{-ia\pi}}|s|^{-1}e^{-i\pi}e^{-s}ds\\ & = & -\int_{0}^{x}\left|\frac{s}{x}\right|^{a}|s|^{-1}e^{-s}ds\\ & > & 0 \end{eqnarray*} hence $g(x)\in\mathbb{R}$.
Then $$ g''(x)=\frac{h''(x)}{h(x)}-\frac{(h'(x))^{2}}{h(x)^{2}}+\frac{a}{x^{2}} $$
Since $h'(x)=x^{a-1}e^{-x}$, we can show $$ h''(x) = \left(\frac{a-1}{x}-1\right)h'(x) $$
Hence \begin{eqnarray*} g''(x) & = & \frac{h''(x)}{h(x)}-\frac{(h'(x))^{2}}{h(x)^{2}}+\frac{a}{x^{2}}\\ & = & \left(\frac{a-1}{x}-1\right)\frac{h'(x)}{h(x)}-\left(\frac{h'(x)}{h(x)}\right)^{2}+\frac{a}{x^{2}} \end{eqnarray*}
We can write the ratio $\frac{h'(x)}{h(x)}$ as follows: \begin{eqnarray*} \frac{h'(x)}{h(x)} & = & \frac{x^{a-1}e^{-x}}{\int_{0}^{x}t^{a-1}e^{-t}dt}\\ & = & \frac{1}{\int_{0}^{x}\left(\frac{t}{x}\right)^{a-1}e^{-t}e^{x}dt}\\ & = & \frac{1}{xe^{x}\int_{0}^{1}s^{a-1}e^{-sx}ds}\\ & = & \frac{1}{xe^{x}\phi(x)} \end{eqnarray*} where $\phi:\mathbb{R}\to\mathbb{R}$ is given by $$ \phi(x)=\int_{0}^{1}s^{a-1}e^{-sx}ds $$
(this also works when $x<0$)
Thus, \begin{eqnarray*} g''(x) & = & \left(\frac{a-1}{x}-1\right)\frac{h'(x)}{h(x)}-\left(\frac{h'(x)}{h(x)}\right)^{2}+\frac{a}{x^{2}}\\ & = & \frac{1}{(xe^{x}\phi(x))^{2}}\left(a(e^{x}\phi(x))^{2}+(a-1-x)e^{x}\phi(x)-1\right) \end{eqnarray*}
First, assume $x\ge0$. Let's consider the cuadratic equation $az^{2}+(a-1-x)z-1=0$. It's roots are given by $$ z_{\pm}(x) = \frac{(x+1-a)}{2a}\pm\frac{\sqrt{(x+1-a)^{2}+4a}}{2a} $$ with $z_{-}(x)\le0,z_{+}(x)>0$.
We can show \begin{eqnarray*} \frac{d}{dx}(e^{x}\phi(x)) & = & e^{x}\int_{0}^{1}(s^{a-1}-s^{a})e^{-sx}ds\\ \frac{d^{2}}{dx^{2}}(e^{x}\phi(x)) & = & e^{x}\int_{0}^{1}(s^{a-1}-2s^{a}+s^{a+1})e^{-sx}ds\\ \frac{dz_{+}}{dx}(x) & = & \frac{1}{2a}\left(1+\frac{(x+1-a)}{\sqrt{(x+1-a)^{2}+4a}}\right)\\ \frac{d^{2}z_{+}}{dx^{2}} & = & \frac{2}{\left((x+1-a)^{2}+4a\right)^{\frac{3}{2}}} \end{eqnarray*} Evaluating the derivatives at 0, we get \begin{eqnarray*} e^{0}\phi(0) & = & \frac{1}{a}\\ \frac{d}{dx}(e^{x}\phi(x))(0) & = & \frac{1}{a}-\frac{1}{a+1}=\frac{1}{a(a+1)}\\ \frac{d^{2}}{dx^{2}}(e^{x}\phi(x))(0) & = & \frac{1}{a}-\frac{2}{a+1}+\frac{1}{a+2}\\ z_{+}(0) & = & \frac{1}{a}\\ \frac{dz_{+}}{dx}(0) & = & \frac{1}{a(a+1)}\\ \frac{d^{2}z_{+}}{dx^{2}}(0) & = & \frac{2}{\left((1-a)^{2}+4a\right)^{\frac{3}{2}}}=\frac{2}{(1+a)^{3}} \end{eqnarray*} We get $\frac{d^{2}}{dx^{2}}(e^{x}\phi(x))\ge\frac{d^{2}z_{+}}{dx^{2}}(x)$ for $x\ge0$, while $e^{0}\phi(0)=z_{+}(0)$ and $\frac{d}{dx}(e^{x}\phi(x))(0)=\frac{dz_{+}}{dx}(0)$. Hence, $e^{x}\phi(x)\ge z_{+}(x)$ when $x\ge0$, hence $g$ convex in such case, from where it follows $f$ is decreasing when $\beta<1$.
For $a<1$, the numerator in $g''$ is less than zero when $x\ll-1$, thus, for such values this function is not convex. Hence, this method does not allow us to prove the initial proposition when $\beta>1$. In such case, we can prove $f$ increasing when $x<\beta^{-1}$.