A new trick for the Sophomore's dream?

Question

Can we try to have for $x \in(\alpha,\beta),\beta>1,\alpha\geq 1$ real numbers:

$$x^{-x}=\sum_{n=1}^{\infty}a_{n}(c_n+x)^{b_{n}}e^{-d_n(c_n+x)^{b_{n}+1}}$$

Where $a_n,b_n,c_n,d_n\in(-\infty,\infty)$

You can find some material in my answer here Prove that $\int_0^\infty\frac1{x^x}\, dx<2$ .

The goal in integrating easily the RHS is to give an infinite series for the integral :

$$\int_{0}^{\infty}x^{-x}dx$$

Using also the Sophomore's dream .

I also tried the Weierstrass's factorization theorem and there is a link with Stirling's series which have explicit representation .

How to find a sequence $a_n,b_n$ ?

Sides notes :

A straight approach using Am-Gm and $a_n$ all equal doesn't work !

We have better result with $a_n$ non equal and for example :

$$f\left(x\right)-x^{-x}<2\cdot10^{-4}$$

Where :

$f\left(x\right)=\frac{\left(ex^{\frac{42}{100}+\frac{12}{1000}}e^{-x^{1+\frac{42}{100}+\frac{12}{1000}}}+\left(1+e^{-1}\right)ex^{\frac{40}{100}+\frac{12}{1000}}e^{-x^{1+\frac{40}{100}+\frac{12}{1000}}}+\left(1-e^{-1}\right)ex^{\frac{41}{100}+\frac{12}{1000}}e^{-x^{1+\frac{41}{100}+\frac{12}{1000}}}\right)}{3}$

Example with the generalized equality :

$$\frac{\left(\frac{e^{d\left(1+c\right)^{a}}}{\left(1+c\right)^{\left(a-1\right)}}\left(x+c\right)^{\left(a-1\right)}e^{-d\left(x+c\right)^{a}}+ex^{0.4212}e^{-x^{1.4212}}\right)}{2}$$

Where : $d=1.4,c=-0.6,a=1.2$

Ps :State like this the problem was impossible on $x\in(1,\infty)$ so I split it .

Answer 1

I put here my progress so far :

We have :

$f\left(x\right)=\frac{\left(ex^{\frac{42}{100}+\frac{12}{1000}}e^{-x^{1+\frac{42}{100}+\frac{12}{1000}}}+\left(1+e^{-1}\right)ex^{\frac{40}{100}+\frac{12}{1000}}e^{-x^{1+\frac{40}{100}+\frac{12}{1000}}}+ex^{\frac{41}{100}+\frac{12}{1000}}e^{-x^{1+\frac{41}{100}+\frac{12}{1000}}}+\left(1-e^{-1}\right)ex^{\frac{41}{100}+\frac{12}{1000}}e^{-x^{1+\frac{41}{100}+\frac{12}{1000}}}\right)}{4}$

Then it seems we have for $x\geq1$ :

$$f\left(x\right)-x^{-x}<9\cdot10^{-5}$$

With the absolute value it seems we have for $x\ge 1$ :

$$\left|g\left(x\right)-x^{-x}\right|<2\cdot10^{-4}$$

$g\left(x\right)=\frac{\left(ex^{\frac{42}{100}+\frac{12}{1000}}e^{-x^{1+\frac{42}{100}+\frac{12}{1000}}}+\left(1+e^{-1}\right)ex^{\frac{40-\frac{119}{1000}}{100}+\frac{12}{1000}}e^{-x^{1+\frac{40-\frac{119}{1000}}{100}+\frac{12}{1000}}}+\left(1-e^{-1}+1\right)ex^{\frac{41}{100}+\frac{12}{1000}}e^{-x^{1+\frac{41}{100}+\frac{12}{1000}}}\right)}{4}$

We have also for $x\in[4/3,2]$ :

$$\left|\left(\left(1-j\right)\frac{e^{d\left(1+c\right)^{a}}}{\left(1+c\right)^{\left(a-1\right)}}\left(x+c\right)^{\left(a-1\right)}e^{-d\left(x+c\right)^{a}}+jex^{0.4212}e^{-x^{1.4212}}\right)-x^{-x}\right|<4\cdot10^{-5}$$

Where :

$$d=1.44,c=-0.588,a=1.618,j=0.9986$$

Another attempt 14/01/2023 :

Let $c=0.921$,$k=3$:

$$f\left(x\right)=\frac{c}{k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k}+\frac{1}{k^{2}}\right)}e^{-x^{\left(1+\frac{1}{k}+\frac{1}{k^{2}}\right)}}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{2}}+\frac{1}{k^{4}}\right)}e^{-x^{\left(1+\frac{1}{k^{2}}+\frac{1}{k^{4}}\right)}}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{4}}+\frac{1}{k^{6}}\right)}e^{-x^{\left(1+\frac{1}{k^{4}}+\frac{1}{k^{6}}\right)}}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{6}}+\frac{1}{k^{8}}\right)}e^{-x^{\left(1+\frac{1}{k^{6}}+\frac{1}{k^{8}}\right)}}$$

Then we have for $x\in[1,1.2]$:

$$\left|f\left(x\right)-x^{-x}\right|<2\cdot10^{-5}$$

Edit 15/01/2023 :

We can use a sort of algorithm :

Let :

$$j\left(x\right)=\frac{c}{k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{d}}+\frac{1}{k^{2}}\right)}e^{-x^{\left(1+\frac{1}{k^{d}}+\frac{1}{k^{2}}\right)}}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{2r}}+\frac{1}{k^{4}}\right)}e^{-x^{\left(1+\frac{1}{k^{2r}}+\frac{1}{k^{4}}\right)}}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{4}}+\frac{1}{k^{6}}\right)}e^{-x^{\left(1+\frac{1}{k^{4}}+\frac{1}{k^{6}}\right)}}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k}ex^{\left(\frac{1}{k^{6t}}+\frac{1}{k^{8}}\right)}e^{-x^{\left(1+\frac{1}{k^{6t}}+\frac{1}{k^{8}}\right)}}$$

Where :

$$t=0.56,r=1.275,d=1.15,c=1.082,k=3$$

We introduce the function :

$$F(x)=\frac{|j(x)-x^{-x}|}{x}$$

We use the tangent line method :

$$F(x)\simeq h(x)=F(2)(x-2)+F(2)$$

Or :

$$xF(x)\simeq H(x)= F(2)x(x-2)+F(2)x$$

We introduce the function :

$$F_1(x)=\frac{||n(x)-x^{-x}|-H(x)|}{x}$$

And again we use the tangent line method...It seems to converge on $(\alpha,\beta)$ where $\alpha,\beta \in (1,2)$

Some details :

$$F\left(x\right)=\frac{\left|j(x)-x^{-x}\right|}{x}$$

$$a_{1}\left(x\right)=\frac{\left|F\left(x\right)-F\left(2\right)x\left(x-2\right)-xF\left(2\right)\right|}{x}$$

$$a_{2}\left(x\right)=a_{1}\left(2\right)\left(x-2\right)+a_{1}\left(2\right)$$

$$a_{3}\left(x\right)=\frac{\left|a_{1}\left(x\right)-xa_{2}\left(x\right)\right|}{x},a_{4}\left(x\right)=a_{3}\left(2\right)\left(x-2\right)+a_{3}\left(2\right),a_{4}\left(x\right)=a_{3}\left(2\right)\left(x-2\right)+a_{3}\left(2\right),a_{5}\left(x\right)=\frac{\left|a_{3}\left(x\right)-xa_{4}\left(x\right)\right|}{x}$$

And so on

The accuracy growths by power of two .