Problem :
Show that :
$$\int_{0}^{\infty}\frac{1}{\left(x!\right)!}dx>2$$
I have tried to apply twice the Stirling formula without success .As other remarks the crucial part is for $x\in[0,3]$.
For the Craft work I have tried to show the inequality for $x\in[0,1]$ :
$$x^{-0.1375x^{1.25}}\leq \frac{1}{\left(x!\right)!}$$
But I cannot find a path for the rest of the crucial interval .
As other attempt the function in the integral seems to be concave on $[0,2]$ wich allows us to a linear approximation but how to show it I haven't the skills for that .
Disgression :
We have also :
$$\int_{0}^{\infty}\left(x^{-x}+\frac{1}{\left(x!\right)!}\right)dx<4$$
So how to show it properly ?
Just an idea
Consider $$I=\int_{a-\epsilon}^{a+\epsilon}\frac{dx}{\left(x!\right)!}$$ Expand the integrand as a series around $x=a$ to $O\left((x-a)^3\right)$
This would give $$\frac{3 \,\Gamma (1+\Gamma (a+1))\,I- 6\epsilon}{\Gamma(a+1)\,\epsilon ^3}=$$ $$\psi ^{(0)}(a+1)^2 \left(-H_{\Gamma (a+1)}+\Gamma (a+1) \psi ^{(0)}(\Gamma (a+1)+1)^2-a \Gamma (a) \psi ^{(1)}(\Gamma (a+1)+1)+\gamma \right)-\psi ^{(1)}(a+1) \psi ^{(0)}(\Gamma (a+1)+1)$$
Trying for $a=\frac 1 {10}$, $\epsilon=\frac 1{100}$ this gives $\color{red}{0.020399654}215$ while numerical integration gives $\color{red}{0.020399654197}$
Using instead the simplest $[1,1]$ Padé approximant, for the same conditions, it would give $\color{red}{0.0203996541}54$