Erik Satie
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Empress Elisabeth of Austria did maths .

She is famous for this integral :

$$\int_{0}^{\pi}Si(x)Si(x)dx$$

$$\exists x>0$$ then we have :

$$\frac{\pi^{2}}{6}=1+f\left(1\right)-\sum_{n=1}^{\infty}n!\left(f\left(x\right)-f\left(1\right)\right)^{n}$$

Where : $$f\left(x\right)=\tanh\left(h\left(h\left(x\right)\right)\right),h\left(x\right)=\operatorname{erf}\left(x\right)$$

The late problem is related to the Cauchy's proof with the inequalities :

$$\frac{\pi^{2}}{6}\left(\frac{2m}{2m+1}\right)\left(\frac{2m-1}{2m+1}\right)<\sum_{n=1}^{m}\frac{1}{n^{2}}<\frac{\pi^{2}}{6}\left(\frac{2m}{2m+1}\right)\left(\frac{2m+2}{2m+1}\right)$$

if we replace by a plus in the late problem as :

$$\frac{\pi^{2}}{6}=1+f\left(1\right)+\sum_{n=1}^{\infty}n!\left(f\left(x\right)-f\left(1\right)\right)^{n}$$

Then $$\exists y$$ such that :

$$f(y)=0$$ and $$0=h\left(\frac{\pi^2}{6}-1\right)$$

Where $$h$$ is obtained using the Lagrange's theorem (inversion) for example in the case $$n=10$$ we have :

Where

$$f\left(x\right)=\tanh\left(\operatorname{erf}\left(\operatorname{erf}\left(x\right)\right)\right)$$

Then on the polynomial $$g$$ we can invert it to get $$h$$ in the case $$n=10$$ wich is an approximation .

Now a conjecture for $$y\geq 1$$:

$$e^{y}=^?\lim_{x \to 0}\left(\left|1-\frac{y-1+x!}{1-\frac{1}{1-x!x!!x!!!x!!!!x!!!!!\cdot\cdot\cdot}}\right|\right)^{\frac{1}{x}}$$

Where $$(x!)!=\Gamma(\Gamma(x+1)+1)$$ and so on

Stronger form found 20/10/2022 :

Let $$x\geq 1$$ :

$$\sqrt{2\pi x}e^{\frac{1}{12x}}\left(\frac{x}{e}\right)^{x}\tanh\left(\ln\left(12x\ln^{2}\left(x\right)+1\right)\right)

We can strenghened this and it seems we have for $$x\geq 1$$ :

$$f\left(x\right)=\frac{\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}e^{\frac{1}{12x}}\tanh\left(\ln\left(12x\ln^{2}\left(x\right)+1+x+\ln\left(x\right)+\sqrt{x}+\sum_{n=1}^{10}2^{2+n}\left(x^{\frac{1}{2^{n}}}-1\right)\right)\right)}{x!}

For $$1\leq x\leq 8$$ $$\exists C>0$$ such that :

$$f\left(x\right)=\sqrt{2\pi x}\left(\frac{x}{e}\right)^{x}e^{\frac{1}{12x}}\tanh\left(\ln\left(12x\ln^{2}\left(x\right)+1+x+\ln\left(x\right)+x+C\left(\sum_{n=1}^{100}n2^{\left(n-\frac{\ln^{2}\left(n\right)}{2}\right)}\left(x^{\frac{1}{2^{n}}}-1\right)\right)\right)\right)-x!<0$$

22/10/2022 23/10/2022

28/10/2022 :

Correction the integral with power tower seems equal to : $$\frac{1+\pi}{10}$$

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