An equality about the minimum of the Gamma function .

Question

Working with Geogebra I have found an intriguing thing perhaps it's well-know :

Let consider the equality on $I=(1,2)$ :

$$\operatorname{W}(x^2)=\Gamma(x)$$

In the LHS we have the Lambert's function or $\operatorname{W}$.

It seems that the equality above occurs ($x= 1.4653\cdots$) really near from the minimum of the Gamma function ($x=1.4616\cdots$).It gives us a good approximation of the minimum of the Gamma function wich is convex .

Edit : See https://ysharificalc.wordpress.com/2021/02/08/the-lambert-w-function-two-integrals/

Question :

How to explain this result ?

Is there a deep link between these two function ?

Thanks and apologize for the bad english .

Answer 1

$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt\Rightarrow\Gamma'(x)=\int_0^\infty \ln(t)t^{x-1}e^{-t}\,dt$$ you want $\Gamma'(x)=0$

---

Also note that $W^{-1}(x)=xe^x$ so if we say: $$y=W(x^2)=\Gamma(x)$$ then: $$x^2=ye^y\Rightarrow x=\sqrt{ye^y}$$ so the intersection point is satisfied by: $$y=\Gamma(\sqrt{ye^y})=\int_0^\infty t^{\sqrt{ye^y}-1}e^{-t}\,dt$$ however I do not see a link between this and our $\Gamma'=0$