How to evaluate $ \int_{0}^{1} x^{x^{x^{x^…}}} dx $ Inspired by the fact that $\int_0^1 \frac{1}{x^x}=\sum_{k=1}^\infty \frac{1}{k^k}$ I asked myself wether it is possible to evaluate the following integral:
$$
\int_{0}^{1} x^{x^{x^{x^…}}} dx
$$
In a similar way. Any help is highly appreciated.
 A: This is not a proper answer. But an approach that explores the recursive aspects of the problem that I imagine lead to the expected solution. Define the functions $\varphi: (0,+\infty)\to [0,+\infty]$ and $\varphi_n: (0,+\infty)\to [0,+\infty]$ by

More precisely these functions are defined by recursion:
$$
\varphi_{n}(x)=
\left\{
\begin{array}{rcl}
x, & \mbox{ if } & n=1\\
x^{\varphi_{n-1}(x)}, & \mbox{ if } & n> 1
\end{array}
\right.
\quad
\mathrm{ and } 
\quad
\varphi(x)=\lim_{n\to \infty}\varphi_n(x)
$$
if the limit exists. In the case of $x$ between $0$ and $1$ such limit exists, i.e. for all $\frac{1}{2}>\epsilon>0$ and $x\in[\epsilon, 1]$ the limit $\varphi(x)=\lim_{n\to\infty}\varphi_n(x)<\infty$ exists. Taking care with convergence limits, 
$$
\int_0^1\varphi(x)\mathrm{d} x =
\lim_{\epsilon\to 0}\int_\epsilon^{1}\lim_{n\to\infty}\varphi_n(x)\mathrm{d} x
$$
By the Lebesgue's Dominated Convergence Theorem ( $|\varphi_n(x)|\leq 1$ for all $x\in[\epsilon, 1]$ for all $\epsilon >0$).
$$
\int_0^1\varphi(x)\mathrm{d} x=
\lim_{\epsilon\to 0}\lim_{n\to\infty}\int_\epsilon^{1}\varphi_n(x)\mathrm{d} x
$$
Now let's get a more tractable procedure to approximate the integral 
$
\int_\epsilon^{1}\varphi_n(x)\mathrm{d} x \quad \mbox{ for all } n.
$
Note that
\begin{align}
 \log\varphi_{n}(x)=&  \varphi_{n-1}(x) \log x \\
 \log^{2}\varphi_{n}(x)=&  \varphi_{n-2}(x) (\log x )^2\\
 \log^{3}\varphi_{n}(x)=&  \varphi_{n-3}(x) (\log x )^3\\
\vdots\;& \hspace{2cm}\vdots \\
\log^{k}\varphi_{n}(x)=&  \varphi_{n-k}(x) (\log x )^k\\
\vdots\;& \hspace{2cm}\vdots \\
\log^{n-2}\varphi_{n}(x)=&  \varphi_{2}(x) (\log x )^{n-2}\\
\log^{n-1}\varphi_{n}(x)=&  \varphi_{1}(x) (\log x )^{n-1}\\
\end{align} 
Then $\varphi_n(x)=\exp^{n-1}\left(x \cdot (\log x )^{n-1}\right)$
\begin{align}
\int_0^{1}\varphi (x)\mathrm{d} x 
&
=
\lim_{\epsilon\to 0}\int_{\epsilon}^{1}\lim_{n\to \infty}\varphi_n(x)\mathrm{d}x
\\
&
=
\lim_{\epsilon\to 0}\int_{\epsilon}^{1}\lim_{n\to \infty}\exp^{n-1}\left(x\cdot (\log x)^{n-1}\right)\mathrm{d}x
\\
&
=
\lim_{\epsilon\to 0}\lim_{n\to \infty}\int_{\epsilon}^{1}\exp^{n-1}\left(x\cdot (\log x)^{n-1}\right)\mathrm{d}x
\\
&
=
\lim_{n\to \infty}\lim_{\epsilon\to 0}\int_{\epsilon}^{1}\exp^{n-1}\left(x\cdot (\log x)^{n-1}\right)\mathrm{d}x
\end{align}
