How to prove monotonicity in this case? 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.
 A: 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 $x<\beta^{-1}$.
