TITLE: Show that $\int\limits_0^1 \left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{\left(x^{x}\right)^{⋰}}}}\ \mathrm{d}x=\frac{\pi^2}{12}$.
QUESTION [18 upvotes]: How can it be shown that
  $$\lim_{p\to\infty}I(p)= \lim_{p \to \infty}\int^{1}_0 (x^x)^{\scriptscriptstyle {(x^x)^{(x^x)^{(x^x)^{(x^x)^{(x^x)...(p \; times)}}}}}} dx= \frac{\pi^2}{12}$$ 

$I(1)=\int^{1}_0 x^x dx=\sum_{n=0} ^{\infty} \frac{1}{n!}\int^{1}_0 \ln(x)^nx^n dx$ 
also $\int^{1}_0 \ln(x)^nx^n dx=(-1)^n (1+n)^{-1-n}n!$ .
I don't know how to calculate $I(p)$ beyond $p=1.$
Note: This integral was proposed on Romanian Mathematical Magazine and two solutions can be found here.

REPLY [16 votes]: From here:
$$W(x)=\sum _{n=1}^{\infty } \frac{(-n)^{n-1} x^n}{n!}$$
and with CAS help:
$$\begin{align}
\int_0^1 -\frac{W(-x \ln (x))}{x \ln (x)} \, dx &=\int_0^1 \left(\sum _{n=1}^{\infty }
   \frac{(-1)^{2 n} n^{-1+n} x^{-1+n} \ln ^{-1+n}(x)}{n!}\right) \, dx \\ 
&=\sum _{n=1}^{\infty }
   \int_0^1 \frac{(-1)^{2 n} n^{-1+n} x^{-1+n} \ln ^{-1+n}(x)}{n!} \, dx\\
&=\sum _{n=1}^{\infty } \frac{\left((-1)^{2 n} n^{-1+n}\right) \int_0^1 x^{-1+n} \ln ^{-1+n}(x) \,
   dx}{n!}\\
&=\sum _{n=1}^{\infty } \frac{\left((-1)^{2 n} n^{-1+n}\right)
   \left(-\left(-\frac{1}{n}\right)^n \Gamma (n)\right)}{n!}\\
&=\sum _{n=1}^{\infty }
   \frac{(-1)^{1+3 n}}{n^2}\\
&=\frac{\pi ^2}{12}.
\end{align}$$