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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.

Addendum:

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:

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

Tolaso
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    "we all know" ? of course no, but if you write the sketch proof maybe it will become obvious – reuns Sep 28 '16 at 12:24
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    and most of what is possible to do with $Li_2(z)$ is probably there – reuns Sep 28 '16 at 12:36
  • 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 $\rm Li_2(\frac{1}{1-x})$. – user90369 Sep 28 '16 at 15:22
  • 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. :) – Tolaso Sep 28 '16 at 17:26
  • Hmm.. I found this (http://math.stackexchange.com/questions/1424600/conjecture-re-operatornameli-2-left-frac12-frac-i6-right-frac7-pi2?rq=1). No hopes of reducing it to something known. :( – Tolaso Sep 28 '16 at 17:46
  • @ Tolaso you should prove $Li_2(z)+Li_2(1-z) = \zeta(2) - \log \log(1-z)$ instead. You say it is very well-known but it seems obvious you don't know where it comes from. (@user90369) – reuns Sep 28 '16 at 19:02
  • @user1952009 I know the relation. Besides it is a fundemental one for the dilog so I recall it by heart. Anyway , there is no hope of actually getting a closed form so I am moving on. I think the approach I outlined (also contained in the link) is enough than good to actually extract the real part. Thanks for participating in the discussion.

    T.

     – Tolaso Sep 28 '16 at 19:49
  • @Tolaso look at a proof of $Li_2(z)+Li_2(1-z)=\ldots$. – reuns Sep 28 '16 at 20:27
  • My evaluation is wrong (sorry), please don't accept it any more so that I can delete it. Thanks ! – user90369 Nov 11 '16 at 06:35
  • @user90369 I deleted my acceptence. – Tolaso Nov 11 '16 at 08:15

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