TITLE: Closed-form of $\int_0^1 \int_0^1 \int_0^1 x^{(y^z)} \,dz\,dy\,dx$
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.
$$ γ_1\left(a,b\right)=\lim _{n\to \infty }\left(-\frac{1}{2}ln\left(n\right)^2+\sum _{k=1}^n\:\frac{ln\left(k+a\right)}{k+b}\right) $$
$$ I=1+2γln\left(2\right)-\frac{1}{2}ln\left(2\right)^2-γ_1\left(1,0\right)+γ_1\left(\frac{1}{2},0\right) $$