Find $\int_a^b(f(x)-g(x))\mathrm{d}x$ if $f(x)=x^{x^{1-x}}$ and $g(x)$ is the linear equation from the extrema $(a,f(a))$, $(b,f(b))$ of $f(x)∀x≥0$.

Question

Find $\int_a^b(f(x)-g(x))\mathrm{d}x|$ if $f(x)=x^{x^{1-x}}$ and $g(x)$ is the linear equation from the extrema $(a,f(a))$, $(b,f(b))$ of $f(x)∀x≥0$.

Find the exact area. The area under $g(x)$ as negative and the area above $g(x)$ is positive.

How I (tried to) solve it

1. I found the x-coordinates of the extrema using $f'(x)=-x^{x^{1-x}-x}(x\ln(x)-1)(\ln(x)+1)=0$; $x=a,b=e^{-1},e^\Omega$ where $\Omega$ represents the omega constant https://en.wikipedia.org/wiki/Omega_constant.
2. I plugged the x-coordinates into $f(x)$ to get the extrema. I got $(e^{-1},e^{-e^{e^{-1}-1}})$ and $(e^{\Omega},e^{e^{-1}})$.
3. I then got $g(x)$ with the two points; $g\left(x\right)=\frac{e^{1+e^{-1}}-e^{1-e^{e^{-1}-1}}}{e^{\Omega+1}-1}x-\frac{e^{e^{-1}}-e^{1-e^{e^{-1}-1}+\Omega}}{e^{1+\Omega}-1}$. Note that I simplified it a lot to get it down to this.
4. To find the area, I did

$\int_a^b(f(x)-g(x))\mathrm{d}x=\int_{e^{-1}}^{e^\Omega}(f(x)-g(x))\mathrm{d}x=F(e^{\Omega})-G(e^{\Omega})-(F(e^{-1})-G(e^{-1}))$ where $F$ and $G$ represent the antiderivatives of their respective functions. Unfortunately, here is where I got stuck because I could not find the antiderivative of $f(x)$ (as I am only in Calc AB). I did, however, put it into Desmos and got the value $0.0369975757242$. What is the exact value?

Desmos "solution": https://www.desmos.com/calculator/yzzhqdptb4

Also, I know this might be a little confusing to read so editing is greatly appreciated if necessary. Thanks!

Answer 1

If you make a series expansion around $x=1$, you have $$f(x)-g(x)=\sum_{n=0}^\infty \alpha_n\,(x-1)^n$$ where $$\alpha_0=\frac{e^{1-e^{\frac{1}{e}-1}} (\Omega-1)-\Omega-(e-1) e^{\frac{1}{e}} \Omega+e}{e-\Omega}$$ $$\alpha_1=1+\frac{e \left(e^{-e^{\frac{1}{e}-1}}-e^{\frac{1}{e}}\right) \Omega}{e-\Omega}$$ and the next coefficients form the sequence $$\left\{0,-1,0,\frac{7}{12},\frac{1}{6},-\frac{169}{360},-\frac{29}{72 0},\frac{589}{2520},\frac{253}{5040},-\frac{11959}{75600},-\frac{131}{302400},\cdots\right\}$$ $$\int_{\frac 1 e}^{\frac 1 \Omega} \big[f(x)-g(x)\big]\,dx=\sum_{n=0}^\infty \alpha_n\,\frac{\left(\frac{1}{\Omega }-1\right)^{n+1}-\left(\frac{1}{e}-1\right)^{n+1}}{n+1}$$

I only gave the first twelve coefficients but the next are easy to generate. Computing the partial sums $$\left( \begin{array}{cc} k & \sum_{n=0}^k \\ 10 & 0.0373722254885034 \\ 20 & 0.0369975152106470 \\ 30 & 0.0369975783245838 \\ 40 & 0.0369975758084201 \\ 50 & 0.0369975757271674 \\ 60 & 0.0369975757242847 \\ 70 & 0.0369975757241703 \\ 80 & 0.0369975757241654 \\ 90 & 0.0369975757241652 \\ 100 & 0.0369975757241652 \\ \end{array} \right)$$

With more decimals, the integral is $$0.036997575724165175495421514762496185924469669290735\cdots$$ which not recognized by the $ISC$. But, what is proposed as a "close closed" form is $$\color{blue}{\large\frac \Omega {10} \, K_1(1){}^{\sin (1)}}$$ which shows an absolute error of $4.486\times 10^{-9}$.