# 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.

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.

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}