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 $\beta>1$.


By drawing the graph for some values with mathematica we can expect that the result is true. Also the sign of derivative is more delicate.


Maybe one would have a smart idea to do it.



How to prove monotonicity of such function?
$\newcommand{\Ga}{\Gamma}$The desired result actually holds for all real $a>0$ and, moreover, $f$ is monotonic on the entire interval $(0,\infty)$. 


Indeed, let $c:=-\alpha>0$ and $b:=\beta>0$. Note that for any real $u\ne0$ we have 
\begin{equation}
 u^{-a}(\Ga(a)-\Ga(a,u))=u^{-a}\int\_0^u t^{a-1}e^{-t}\,dt
 =\int\_0^1 s^{a-1}e^{-us}\,ds. 
\end{equation}
So, we may let $0^{-a}(\Ga(a)-\Ga(a,0)):=\int\_0^1 s^{a-1}e^{-0s}\,ds=1/a$, by continuity. 


So, we may write 
\begin{equation}
 f(x)=J\_b(x)/J\_1(x),
\end{equation}
where 
\begin{equation}
 J\_b(x):=\int\_0^1 s^{a-1}x^{cs}b^{cs}\,ds.
\end{equation}


Hence, $f'(x)$ is equal in sign to 
\begin{align\*}
& \tfrac2c\,J'\_b(x)J\_1(x)-\tfrac2c\,J'\_1(x)J\_b(x) \\ 
& =\int\_0^1 s^a x^{cs-1}b^{cs}\,ds\, \int\_0^1 t^{a-1}x^{ct}\,dt \\ 
& +\int\_0^1 t^a x^{ct-1}b^{ct}\,dt\, \int\_0^1 s^{a-1}x^{cs}\,ds \\ 
&-\int\_0^1 s^a x^{cs-1}\,ds\, \int\_0^1 t^{a-1}x^{ct}b^{ct}\,dt \\ 
&-\int\_0^1 t^a x^{ct-1}\,dt\, \int\_0^1 s^{a-1}x^{cs}b^{cs}\,ds \\ 
&=\int\_0^1\int\_0^1 ds\,dt\,x^{cs+ct-1}(st)^{a-1}(sb^{cs}+tb^{ct}-sb^{ct}-tb^{cs}) \\ 
&=\int\_0^1\int\_0^1 ds\,dt\,x^{cs+ct-1}(st)^{a-1}(s-t)(b^{cs}-b^{ct}). 
\end{align\*}
Now the result follows, because $(s-t)(b^{cs}-b^{ct})$ is equal in sign to $b-1$ for all distinct real $s$ and $t$. 


