Hi It's a problem create by me :
$$\int_{\frac{1}{2}}^{1}x^x\operatorname{W(x)}\operatorname{dx}<\frac{1}{9c-15}≈0.193990261767$$ Where $c$ is the Takeuchi-Prellberg Constant
I think it's interesting to have the Lambert's function in the expression of the Takeuchi-Prellberg Constant, in the asymptotic behavior of the Bell's numbers and in the integral .Furthermore as we know $x^x=z$ have as solution $x=\exp(\operatorname{W(ln(z))})$
What I have tried so far :
I use power series of $f(x)=x^x$ at $x=1$ we have :
$$f(x)=1 + (x - 1) + (x - 1)^2 + 1/2 (x - 1)^3 + 1/3 (x - 1)^4 + 1/12 (x - 1)^5 + 3/40 (x - 1)^6 - 1/120 (x - 1)^7 + (59 (x - 1)^8)/2520 +O((x - 1)^9) $$
And :
$$g(x)=\operatorname{W_0(x)}=\sum_{n=1}^{\infty}\frac{(-n)^{n-1}}{n!}x^n$$
Remains to multiply the two troncated series and integrate to get the desired result
My question :
Can someone improve the upper bound or give a closed form ?
Thanks a lot for your time and patience
Developing the integrand as a series around $x=1$, we have $$x^x\,W(x)=w+\frac{w (w+2)}{(w+1)}(x-1)+\frac{w (w+2) (2 w^2+3w+2) }{2 (w+1)^3}(x-1)^2+\frac{w \left(w^2+3 w+3\right) \left(3 w^3+11 w^2+8 w+3\right)}{6 (w+1)^5}(x-1)^3+\frac{w \left(8 w^7+58 w^6+180 w^5+321 w^4+368 w^3+276 w^2+104 w+20\right) }{24 (w+1)^7 }(x-1)^4+O\left((x-1)^5\right)$$ where $w=W(1)=\Omega$.
As a function of $n$ the dgeree of expansion, we then have for the integral a closed form $$I_n= \frac{w \,P_{2n-1}(w)}{2^{n+1}\, (n+1)!\,(1+w)^{2n-1}}$$ the numerical results are $$\left( \begin{array}{cc} n & \text{value} \\ 0 & 0.283571645204892 \\ 1 & 0.167441701824308 \\ 2 & 0.201682086153625 \\ 3 & 0.192560070537807 \\ 4 & 0.194346190147895 \\ 5 & 0.193957003195521 \\ 6 & 0.194004358724321 \\ 7 & 0.193991317451300 \\ 8 & 0.193991160551488 \\ 9 & 0.193990511889642 \\ 10 & 0.193990358194065 \\ 11 & 0.193990294268921 \\ 12 & 0.193990272085633 \\ 13 & 0.193990263830734 \\ 14 & 0.193990260767764 \\ 15 & 0.193990259611595 \\ 16 & 0.193990259170115 \\ 17 & 0.193990258999433 \\ 18 & 0.193990258932675 \\ 19 & 0.193990258906272 \\ 20 & 0.193990258895718 \\ \cdots & \cdots \\ \infty &0.193990258888496 \end{array} \right)$$
Edit
For the fun of it (this comes from a friend of mine) an upper bound of the ntegral is $$\frac{-2+2 \sqrt{3}-3 e+6 \pi -2 \pi ^2+7 \log (2)}{5+11 \sqrt{2}+2 \sqrt{3}-3 e-\pi -2 \pi ^2-7 \log (2)-2 \log (3)}$$ for a relative error of $1.44\times 10^{-17}$%. Looking more serious is (from the inverse symbolic calculator) $$\frac{\sqrt[4]{3} \left(16-\sqrt[3]{2}\right)}{100} $$ It also looks very similar to the smallest root of$$1928 x^2-4833 x+865=0\implies x=\frac{4833-\sqrt{16687009}}{3856}$$