14
$\begingroup$

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 ?

$\endgroup$
2
  • 2
    $\begingroup$ If you make it for $x\in[0,e]$, you win $\endgroup$ Aug 2, 2022 at 13:25
  • $\begingroup$ OOF, it's close : apparently Wolfram says the integral is $\sim 2.00356$, dead close to $2$, eh?!? $\endgroup$ Aug 2, 2022 at 15:27

1 Answer 1

4
$\begingroup$

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 $\color{red}{0.0203996541}54$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .