Closed form for ${\large\int}_0^1x\,\operatorname{li}\!\left(\frac1x\right)\ln^{1/4}\!\left(\frac1x\right)dx$ Let $\operatorname{li}(x)$ denote the logarithmic integral:
$$\operatorname{li}(x)=\int_0^x\frac{dt}{\ln t}.$$
How can we prove the following conjectured closed form?
$${\large\int}_0^1x\,\operatorname{li}\!\left(\frac1x\right)\ln^{1/4}\!\left(\frac1x\right)dx\stackrel{\color{gray}?}=\left(\frac12-\frac{\operatorname{arctan}\left(\sqrt[4]2\right)+\operatorname{arcoth}\left(\sqrt[4]2\right)}{4\sqrt[4]2}\right)\cdot\Gamma\!\left(\frac14\right)$$
Related questions: [1][2].
 A: Let $x = e^{-y}$ and denote the value of the integral by $I$. Then we have
\begin{equation}
I = \int_0^\infty y^{1/4} e^{-2y} \text{ li}(e^y) \, dy.
\end{equation}
Consider the parameter
\begin{equation}
I(b) = \int_0^\infty y^{1/4} e^{-2y} \text{ li}(e^{by}) \, dy.
\end{equation}
Differentiating we obtain
\begin{equation}
I'(b) = \frac{1}{b} \int_0^\infty y^{1/4} e^{-(2-b)y} \, dy.
\end{equation}
Letting $u = (2-b)y$, we obtain
\begin{equation}
I'(b) = \frac{\Gamma\left(\frac{1}{4} \right)}{4 b(2-b)^{5/4}},
\end{equation}
where we require $b < 2$ for convergence of the integral. Using mathematica to evaluate the integral we obtain
\begin{equation}
I(b) = \frac{\Gamma\left(\frac{1}{4} \right)}{4} \left \{\frac{2}{(2-b)^{1/4}} + \frac{\arctan \left[ \left(1 - \frac{b}{2} \right) \right]^{1/4} - \text{arctanh} \left[ \left(1 - \frac{b}{2} \right) \right]^{1/4}}{2^{1/4}} \right \} + C.
\end{equation}
The constant of integration $C$ can be determined by letting $b \to -\infty$ which gives
\begin{equation}
C = - \frac{\pi \Gamma \left(\frac{1}{4}\right) e^{i \pi/4}}{4 \cdot 2^{3/4}}.
\end{equation}
Finally, if we note $I = I(1)$ we obtain
\begin{equation}
I(1) = \Gamma\left(\frac{1}{4} \right) \left [ \frac{1}{2} + \frac{\arctan \left( 2^{-1/4} \right) - \text{arctanh} \left( 2^{-1/4} \right)}{4 \cdot 2^{1/4}} - \frac{\pi e^{i \pi/4}}{4 \cdot 2^{3/4}} \right ].
\end{equation}
Taking the real part results in the correct numerical value when plugged into mathematica (as does your result).
