Closed-form of $\int_0^1 \int_0^1 \int_0^1 x^{(y^z)} \,dz\,dy\,dx$ We know that
$$\int_0^1 \int_0^1 x^y\,dy\,dx = \ln 2.$$
Do we know a closed-form of
$$\int_0^1 \int_0^1 \int_0^1 x^{(y^z)} \,dz\,dy\,dx\,?$$
As a start we know that
$$\int_0^1 x^{(y^z)}\,dz = \frac{\operatorname{Ei}(y \ln x) - \operatorname{li}(x)}{\ln y}.$$
Here $\operatorname{Ei}$ is the exponential integral, and $\operatorname{li}$ is the logarithmic integral.
 A: First integrate with respect to $x$ i.e
$$\int_0^1 \int_0^1 \int_0^1 x^{y^z}\,dx\,dy\,dz=\int_0^1 \int_0^1 \frac{1}{1+y^z}\,dy\,dz$$
Since $0<y<1$ and $0<z<1$,
$$\int_0^1 \int_0^1 \frac{1}{1+y^z}\,dy\,dz=\sum_{k=0}^{\infty} (-1)^k \int_0^1 \int_0^1 y^{kz}\,dy\,dz=\sum_{k=0}^{\infty} \frac{(-1)^k\ln(1+k)}{k}$$
$$\begin{aligned}
\sum_{k=0}^{\infty} \frac{(-1)^k\ln(1+k)}{k} & =1+\sum_{k=1}^{\infty} \frac{(-1)^k\ln(1+k)}{k} \\
&=1-\sum_{k=1}^{\infty} \frac{(-1)^{k-1}\ln k}{k}+\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\ln\left(1+\frac{1}{k}\right) \\
\end{aligned}$$
The first sum is $-\eta'(1)$ where $\eta(s)$ is dirichlet eta function.
For the second sum:
$$\begin{aligned}
\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\ln\left(1+\frac{1}{k}\right) &=-\sum_{k=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^k}{k}\frac{(-1)^{m}}{mk^m} \\
&=\sum_{m=1}^{\infty} \frac{(-1)^m}{m}(1-2^{-m})\zeta(m+1)\\
&=\sum_{m=1}^{\infty} \frac{(-1)^m}{m}\zeta(m+1)-\sum_{m=1}^{\infty} \left(\frac{-1}{2}\right)^{m}\frac{\zeta(m+1)}{m} \\
\end{aligned}$$
I am thinking of manipulating the following:
$$\sum_{n=1}^{\infty} \zeta(n+1)x^n=-\gamma-\psi(1-x)$$
but I am not sure how to do that.
Edit: As Varun in the comments pointed out, we could use
$$\sum_{n=1}^{\infty} \frac{\zeta(n+1)}{n}x^n = \int_{0}^{x}\frac{-\gamma-\psi(1-t)}{t}dt$$
Using this and after some simplification, the second sum simplifies to,
$$-\gamma\ln 2 + \int_{1/2}^{1}\frac{\psi(1+t)}{t}dt = -\gamma\ln 2 -\int_{1/2}^{1} \frac{\psi(t)}{t} dt - 1$$
where I used $\psi(1+t) = \psi(t) + 1/t$.
With $\eta'(1) = \gamma\ln 2 - \frac{\ln^2 2}{2}$, we end up with what was posted by another user in the thread,
$$-\int_{1/2}^1 \frac{\psi(t)}{t}dt + \frac{1}{2}\ln^22$$
Not sure if this leads anywhere but I think this form is interesting nonetheless.
A: I found the answer while browsing Reddit and thought you might want to hear about it. although, I can't really prove it on my own. Edit: I managed to find a way to compute it.
$$\int_0^1 \int_0^1 \int_0^1 x^{y^z}\,dx\,dy\,dz= - \int_ {1/2}^{1} \frac{ψ(x)}{x} \,  dx - \frac{1}{2} \, \ln^{2}(2). $$
This is not a coincidence and I can prove it.
$$I=\int_0^1 \int_0^1 \int_0^1 x^{y^z}\,dx\,dy\,dz = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{∞}y^{nz}\ln\left(x\right)^{n}dxdydz 
= 1-\sum_{n=1}^{∞}\left(-1\right)^{n-1}\ \frac{\ln\left(n+1\right)}{n}$$
$$ \int_{0}^{1}x^{n-1}li\left(x\right)dx=-\frac{\ln\left(n+1\right)}{n} (observation)$$
$$ I = 1+\int_{0}^{1}\frac{li\left(x\right)}{x+1}dx $$
$$I(t) = 1+\int_{0}^{1}\frac{li\left(x^t\right)}{x+1}dx , I(1)=I, I(∞)=1$$
$$ I'(t)=\frac{1}{t}\int_{0}^{1}\frac{x^{t}}{x+1}dx = \frac{\ln2+ψ\left(\frac{t}{2}+1\right)-ψ\left(t+1\right)}{t}=\frac{\ln2+ψ\left(\frac{t}{2}\right)-ψ\left(t\right)}{t}+\frac{1}{t^{2}}  $$
$$ I= 1-\int_{1}^{∞}\left(\frac{\ln2+ψ\left(\frac{t}{2}\right)-ψ\left(t\right)}{t}+\frac{1}{t^{2}}\right)dt = -\int_{1}^{∞}\frac{\ln\left(2\right)+ψ\left(\frac{t}{2}\right)-ψ\left(t\right)}{t}dt  $$
$$  F\left(t\right)=\int_{1}^{t}\frac{ψ\left(x\right)}{x}dx≃\frac{1}{2}\left(ψ\left(\frac{t}{2}+1\right)^{2}-ψ\left(t+1\right)^{2}\right)  ,F(1)=0 $$
$$ -\lim _{t\to \infty }\left(ln\left(2\right)ln\left(t\right)+F\left(\frac{t}{2}\right)-F\left(t\right)\right)+F\left(\frac{1}{2}\right)-F\left(1\right) $$
$$ (\lim _{t\to \infty }\left(ψ\left(\frac{t}{2}+1\right)-ψ\left(t+1\right)\right)=-ln\left(2\right)) $$
$$ =-\lim _{t\to \infty }\left(\ln \left(2\right)\ln \:t+\frac{1}{2}\left(ψ\left(\frac{t}{2}+1\right)^2-ψ\left(t+1\right)^2\right)\right)$$
$$ =-\lim _{t\to \infty }\left(\ln \:t-\frac{1}{2}ψ\left(\frac{t}{2}+1\right)-\frac{1}{2}ψ\left(t+1\right)\right)ln\left(2\right) $$
$$ =-\lim _{t\to \infty }\left(\frac{1}{2}ln\left(\frac{t}{2}\right)+\frac{1}{2}ln\left(2\right)+\frac{1}{2}ln\left(t\right)-\frac{1}{2}ψ\left(\frac{t}{2}+1\right)-\frac{1}{2}ψ\left(t+1\right)\right)ln\left(2\right) $$
$$\lim _{x\to \infty }\left(ln\left(x\right)-ψ\left(x+1\right)\right)=0 $$
$$= -\frac{1}{2}ln\left(2\right)^2+F\left(\frac{1}{2}\right) = -\frac{1}{2}ln\left(2\right)^2-\int _{\frac{1}{2}}^1\:\frac{ψ\left(x\right)}{x}dx$$
Edit 2: The answer can also be written in terms of generalized Stieltjes constant.
$$ γ_1\left(a,b\right)=\lim _{n\to \infty }\left(-\frac{1}{2}ln\left(n\right)^2+\sum _{k=1}^n\:\frac{ln\left(k+a\right)}{k+b}\right) $$
$$ I=1+2γln\left(2\right)-\frac{1}{2}ln\left(2\right)^2-γ_1\left(1,0\right)+γ_1\left(\frac{1}{2},0\right) $$
A: It seems that it does not have a closed-form...

Try $z$
$$\begin{aligned}
  {I_{zyx}} &= \iiint\limits_{{{[0,1]}^3}} {{x^{{y^z}}}{\text{d}}z\;{\text{d}}y\;{\text{d}}x}\\
   &= \iint\limits_{{{[0,1]}^2}} {\frac{{{\text{Ei}}(y\log (x)) - {\text{li}}(x)}}{{\log (y)}}{\text{d}}y\;{\text{d}}x}\\
   &= \int_0^1 {\left( {{{\log }_y}(2) - {{\log }_y}\left( {\frac{{y + 1}}{y}} \right)} \right){\text{d}}y} \quad \log y \mapsto r\\
   &= \int_{ - \infty }^0 {\frac{{{e^r}}}{r}\left( {r - \log \left( {{e^r} + 1} \right) + \log (2)} \right){\text{d}}r}
\end{aligned}$$

Try $y$
$$\begin{aligned}
  {I_{yxz}} &= \iiint\limits_{{{[0,1]}^3}} {{x^{{y^z}}}{\text{d}}y\;{\text{d}}x\;{\text{d}}z}\\
   &= \iint\limits_{{{[0,1]}^2}} {{{(\log (x) + 1)}^{1 - z}}\Gamma \left( {\frac{1}{z}, - \log (x)} \right)\frac{{{\text{d}}x\;{\text{d}}z}}{z}}
\end{aligned}$$

Try $x$
$$\begin{aligned}
  {I_{xyz}} &= \iiint\limits_{{{[0,1]}^3}} {{x^{{y^z}}}{\text{d}}x\;{\text{d}}y\;{\text{d}}z}\\
   &= \iint\limits_{{{[0,1]}^2}} {\frac{1}{{{y^z} + 1}}{\text{d}}y\;{\text{d}}z}\\
   &= \int_0^1 {\frac{{{\psi ^{(0)}}\left( {\frac{{z + 1}}{{2z}}} \right) - {\psi ^{(0)}}\left( {\frac{1}{{2z}}} \right)}}{{2z}}} {\mkern 1mu} {\text{d}}z\quad z \mapsto 1/r\\
   &= \int_0^\infty  {\frac{{{\psi ^{(0)}}\left( {\frac{r}{2}} \right) - {\psi ^{(0)}}\left( {\frac{{r + 1}}{2}} \right)}}{{2r}}} {\mkern 1mu} {\text{d}}r 
\end{aligned}$$
