Prove the convergence of the integral $\int_{0}^{\infty}\frac{x^n}{\Gamma\Big(\operatorname{W}(x)\Big)}dx$

Question

Prove that the following integral is convergent $\forall \,n\geq 1$ a natural number :

$$\int_{0}^{\infty}\frac{x^n}{\Gamma\Big(\operatorname{W}(x)\Big)}dx$$

Where we have the Gamma function and the Lambert's function .

I have tried to use the harmonic-geometric mean inequality because we have :

$$ \frac{\Gamma(x)}{\Gamma(x+n)}=\frac{1}{x(x+1)\ldots(x+n-1)}\leq \Big(\frac{1}{n}\Big(\frac{1}{x}+\frac{1}{x+1}+\ldots+\frac{1}{x+n-1}\Big)\Big)^n$$

And after use the Digamma function to clear the harmonic sum but I'm stuck because I don't know what to do with the Lambert's function (in the Digamma function) .

Warning this integral increases when $n$ increases

Any idea is appreciated .

Thanks a lot for all your contribution.

Answer 1

Letting $x=ze^z$ we have $$ I_n = \int_{0}^{+\infty}\frac{x^n}{\Gamma(W(x))}\,dx = \int_{0}^{+\infty}\frac{(z+1)z^{n+1} e^{(n+1)z}}{\Gamma(z+1)}\,dz $$ and this is simply convergent by Stirling's inequality $$ \Gamma(z+1) \geq \color{red}{z^z} e^{-z} \sqrt{2\pi z}. $$ By the Cauchy-Schwarz inequality the sequence $\{I_n\}_{n\geq 1}$ is not only increasing but also log-convex.
Very roughly $I_n$ behaves like $\exp\exp n$; this can be sharpened through Laplace's method.