I am trying to find if the following inequality is correct or not
\begin{align} f_a(x) \le f_b(x), \forall x\in \mathbb{R} \end{align} for $0<a\le b$, where \begin{align} f_a(x)= \frac{\gamma \left( \frac{1}{a}, \frac{|x|^a}{2} \right)}{\Gamma \left( \frac{1}{a} \right)}, \end{align}
and where the gamma functions are defined as follows \begin{align} \Gamma\left(x \right)= \int_0^\infty t^{x-1} e^{-t} dt, \\ \gamma(x,s) = \int_0^s t^{x-1}\,e^{-t}\,dt. \end{align}
I tried some simulation and this inequality seems to hold. However, I can not show it.
This inequality can also be seen as a monotonicity result in terms of a variable $a$ for $f_a(x)$.
Simulation, results seem to suggest the inequality is true, see the figure below (on the figure x-axis is $x$).
Thanks.
We may assume WLOG $x>0$ and consider that:
$$ f_a(x) = \frac{\int_{0}^{x^a/2}t^{1/a-1}e^{-t}\,dt}{\int_{0}^{+\infty}t^{1/a-1}e^{-t}\,dt}=\frac{\int_{0}^{x}e^{-t^a/2}\,dt}{\int_{0}^{+\infty}e^{-t^a/2}\,dt}$$ so the inequality $f_a\leq f_b$ boils down to: $$ \int_{0}^{x}e^{-t^a/2} dt \int_{0}^{+\infty} e^{-s^b/2}\,ds \leq \int_{0}^{x} e^{-s^b/2}\,ds \int_{0}^{+\infty}e^{-t^a/2}\,dt $$ or to: $$ \int_{0}^{x}\int_{0}^{+\infty}\exp\left(-\frac{t^a+s^b}{2}\right)\,ds\,dt \leq \int_{0}^{x}\int_{0}^{+\infty}\exp\left(-\frac{t^a+s^b}{2}\right)\,dt\,ds $$ that is trivial due to the monotonicity of the argument of the exponential and the transformation $(s,t)\mapsto (t,s)$.