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.
$\newcommand{\Ga}{\Gamma}$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 $t$.
