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: $$\int_0^1 x^x \mathrm{d} x=\sum_0^{\infty} \frac{(-1)^n}{(n+1)^{n+1}}=1-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}+ \dots \approx 0.783430$$ The result for $x^{-x}$ is similar: $$\int_0^1 x^{-x} \mathrm{d} x=\sum_0^{\infty} \frac{1}{(n+1)^{n+1}}=1+\frac{1}{2^2}+\frac{1}{3^3}+\frac{1}{4^4}+ \dots \approx 1.291285$$ 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 $x=\dfrac{1}{e}$ because: $$\begin{align} y&=x^x\\ \frac{\mathrm{d}}{\mathrm{d} x} y &=\frac{\mathrm{d}}{\mathrm{d} x } x^x\\&=\frac{\mathrm{d}}{\mathrm{d} x} e^{\,x \ln x}\\&=e^{\,x \ln x} (1+\ln x)\\&=0 \text { when }\ln x = -1 \text{ or } x=\frac{1}{e} \end{align}$$ 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 $x^x$.