Integration of x^x and x^(-x)

There are many YouTube videos outlining how to integrate $x^x$ and $x^{-x}$ but the following is the best explained that I've come across.

In the video, the function $x^x$ is integrated from 0 to 1. In fact, this proper integral is converted into an infinite series so traditional integration is not possible. The result is quite impressive:$${\int }_0^1{x}^x\mathrm{d}x=\sum _0^{\mathrm{\infty }}\frac{(-1{)}^n}{(n+1{)}^{n+1}}=1-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}+\cdots \approx 0.783430$$The result for $x^{-x}$ is similar:$${\int }_0^1{x}^{-x}\mathrm{d}x=\sum _0^{\mathrm{\infty }}\frac{1}{(n+1{)}^{n+1}}=1+\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^4}+\cdots \approx 1.291285$$Figure 1 shows the SageMath code to draw the two graphs in the given range and to calculate the area bounded by them:

Figure 1: permalink

Figure 2 shows the result with annotations added:

Figure 2

It's hard to see but the area bounded by the two curves is shown at the bottom left and is approximately 0.507855 square units. Figure 3 shows an alternative representation using GeoGebra:

Figure 3

These graphs, their turning points and the bounded area between them are most interesting. Let's not forget that the turning points for both graphs occur when $x={\displaystyle \frac{1}{e}}$ because:$$\begin{array}{rl}y& =x^x\\ \frac{\mathrm{d}}{\mathrm{d}x}y& =\frac{\mathrm{d}}{\mathrm{d}x}{x}^x\\ & =\frac{\mathrm{d}}{\mathrm{d}x}{e}^{x\ln x}\\ & =e^{x\ln x}(1+\ln x)\\ & =0\text{ when }\ln x=-1\text{ or }x=\frac{1}{e}\end{array}$$The result is the same for $y=x^{-x}$, only the y values differ with $e^{1/e}\approx 1.44$ for $x^{-x}$ and $e^{-1/e}\approx 0.69$ for $x^x$.