How to show $a\mapsto \frac{\gamma(a,x)}{\Gamma(a)}$ is decreasing on $\mathbb{R}_+^*$?

Question

Let $a>0,x\geq 0$, the lower regularized incomplete gamma function is defined as : $$P(a,x)=\frac{\gamma(a,x)}{\Gamma(a)} = \int_0^x \frac{e^{-t}t^{a-1}}{\Gamma(a)}dt.$$

I have read in the paper of Gautschi "The incomplete gamma functions since Tricomi" that $P(.,x): a\mapsto \frac{\gamma(a,x)}{\Gamma(a)}$ is decreasing on $\mathbb{R}_+^*$

But how to prove this result ?

Answer 1

It is a little too elementary for this site, but here's the proof: First,

$$ \frac{\partial}{\partial a} P(a,x)=\frac{\Gamma(a)\partial_a\gamma(a,x)-\gamma(a,x)\Gamma'(a)}{\Gamma(a)^2}. $$

So it suffices to show

$$\Gamma(a)\partial_a\gamma(a,x)<\gamma(a,x)\Gamma'(a),$$

or

$$(\Gamma(a)-\gamma(a,x))\partial_a\gamma(a,x)<\gamma(a,x)(\Gamma'(a)-\partial_a\gamma(a,x)),$$

or

$$ \frac{\partial_a\gamma(a,x)}{\gamma(a,x)}<\frac{\Gamma'(a)-\partial_a\gamma(a,x)}{\Gamma(a)-\gamma(a,x)}, $$

or

$$ \frac{\int_0^x e^{-t}t^{a-1}\ln tdt}{\int_0^x e^{-t}t^{a-1}dt}<\frac{\int_x^\infty e^{-t}t^{a-1}\ln tdt}{\int_x^\infty e^{-t}t^{a-1}dt}. $$

Now it is clear that the LHS < $\ln x$ < the RHS.