Evaluate $\int_{0}^{1} (x^x)^{(x^x)^{(x^x)^{^{{\cdot}^{\cdot}}}}} \text{d}x$
This doesn't seem elementary. But since MSE would otherwise want me to show what I have tried, I decided to write it in a more closed form. Write it as $y$, then it becomes $x^{xy} = y \implies x\ln(x) = \frac{\ln(y)}{y}$. Now that $\int \frac{\ln(y)}{y} \text{d}y = \frac{1}{2} \ln^2(y) + c,$ I feel like there's something we can do with this.
Let $W(z)$ denote the principal branch of the Lambert W-function. Denoting the integrand by $y$, we have
$$ y = (x^x)^y = e^{yx \log x}. $$
Multiplying $-x \log x$ to both sides and simplifying,
$$ (-yx \log x) e^{-yx \log x} = -x \log x. $$
Since $0 < y < 1$ and $0 < -x \log x < e^{-1}$ for $0 < x < 1$, we may take $W(\cdot)$ to both sides. This leads to
$$ y = \frac{W(-x \log x)}{-x \log x}. $$
Now by invoking the Taylor series for $W(z)$:
$$ W(z) = \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!} z^n, \qquad |z| < e^{-1}, $$
we have
\begin{align*} \int_{0}^{1} y \, \mathrm{d}x &= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!} \int_{0}^{1} (-x \log x)^{n-1} \, \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!} \int_{0}^{\infty} s^{n-1} e^{-n s} \, \mathrm{d}s, \tag{$x = e^{-s}$} \\ &= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}}{n!} \frac{(n-1)!}{n^n} \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \\ &= \frac{\pi^2}{12} \approx 0.8224670334241132182\cdots. \end{align*}
