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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].

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    $\begingroup$ $I(a) ~=~ \displaystyle\int_0^1x^a~\text{li}\bigg(\frac1x\bigg)~dx ~=~ \frac{\ln a}{a+1}.~$ Now evaluate its fractional derivative of order $\dfrac14.$ $\endgroup$
    – Lucian
    Commented Oct 18, 2015 at 6:26
  • $\begingroup$ Is there a sign error? $\endgroup$
    – Gregory
    Commented Oct 18, 2015 at 6:37

1 Answer 1

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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).

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