We all know that

$$\Re{ \left[{\rm Li}_{2}\left(\frac{1}{2}+iq\right) \right]}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{\left(\frac{1+4q^2}{4}\right)}}^{2}-\frac{{\arctan^2{(2q)}}}{2}$$

where $q \in \mathbb{Q}$. How about the imaginary part of the equation? I have a feeling it involves beta Dirichlet function.


Sketch of proof:

Recall the fact that:

$${\rm Li}_2(\bar{z})=\overline{{\rm Li}_2(z)}$$

hence $$\Re{\rm Li}_2(z)=\frac{{\rm Li}_2(\bar{z}) + \overline{{\rm Li}_2(z)}}{2}$$

and then combine it with the very known functional equation

$${\rm Li}_2(z)+{\rm Li}_2(1-z)=\zeta(2)-\ln z \ln (1-z)$$

Thus the result. Maybe we can get the imaginary part by invoking the known relation:

$$\Im \left[ {\rm Li}_2 \right] =\frac{{\rm Li}_2(\bar{z}) - \overline{{\rm Li}_2(z)}}{2}$$

  • 2
    $\begingroup$ "we all know" ? of course no, but if you write the sketch proof maybe it will become obvious $\endgroup$ – reuns Sep 28 '16 at 12:24
  • 1
    $\begingroup$ and most of what is possible to do with $Li_2(z)$ is probably there $\endgroup$ – reuns Sep 28 '16 at 12:36
  • $\begingroup$ we all know - you make me laugh :-D ; I agree with user1952009 - it would be nice to see where it comes from. Maybe it's senseful to set $\frac{1}{2}+iq=\frac{1}{1-e^{-i2x}}$ because exist formulas with $\rm Li_2(\frac{1}{1-x})$. $\endgroup$ – user90369 Sep 28 '16 at 15:22
  • $\begingroup$ I added a sketch of proof for the "known" fact. I thought it was wide known especially for someone whose hobby is Calculus and Special functions but then again maybe I am mistaken. :) $\endgroup$ – Tolaso Sep 28 '16 at 17:26
  • $\begingroup$ Hmm.. I found this (math.stackexchange.com/questions/1424600/…). No hopes of reducing it to something known. :( $\endgroup$ – Tolaso Sep 28 '16 at 17:46

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