Show an integral inequality involving the golden ratio, $\zeta(2)$ and the gamma function.

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Problem :

Let :

$$\int_{0}^{\infty}\left(\frac{\left(1+\sqrt{5}\right)}{2}\right)^{-x!}x^{2}dx<\frac{\pi^{2}}{6}\frac{\left(1+\sqrt{5}\right)}{2}$$

It's a crude estimation so I think it's doable by hand .

I have tried an intregration by part but nothing comes .

I can show the convergence as :

for $x>0$ we have :

$$x!>x/2$$

Proof :

On $(0,1)$ we have $x!>x>x/2$ and on $(1,\infty)$ we have $f(x)=x!>f'(1)(x-1)+f(1)>x/2$ because the Gamma function is convex .

Wich gives a result around $143.59$

My progress so far :

Using the lemma 7.1 (see reference) we have for $0<x<\frac{\left(1+\sqrt{5}\right)}{2}$:

$$\left(\left(1-x!\right)^{2}+\left(\frac{\left(1+\sqrt{5}\right)}{2}\right)^{-1}x!(2-x!)-\left(\frac{\left(1+\sqrt{5}\right)}{2}\right)^{-1}x!(1-x!)\ln\left(\left(\frac{\left(1+\sqrt{5}\right)}{2}\right)^{-1}\right)\right)x^{2}\geq \left(\frac{\left(1+\sqrt{5}\right)}{2}\right)^{-x!}x^2=f(x)$$

And numerically it seems we have for $12/5<x$ :

$$\frac{f\left(2\right)}{g\left(2\right)}g\left(x+\frac{39}{1000}\right)>\left(\frac{\left(1+\sqrt{5}\right)}{2}\right)^{-x!}x^{2}$$

Where :

$$g\left(x\right)=e^{1-\left(x-2\right)^{e}}$$

How to show it by hand without a computer ?

Reference :

VASILE CIRTOAJE, PROOFS OF THREE OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS, Journal of Nonlinear Sciences and Applications, 4 (2011), no. 2, 130-137