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Throughout this post, $\operatorname{Li}_{2}(x)$ refers to Dilogarithm.

While playing with some Fourier Transforms, I came up with the following expressions:

$$2 \operatorname{Li}_{2}\left(\frac12 \right) + \frac{\pi^{2}}{6} = \sum_{n=1}^{\infty} \left [ \pi (2n-1)^2 \tan^{-1}\left ( \frac{\pi \log(2)}{(2n-1)(2n)\pi^{2}+\log(2)^2}\right ) + \\ \pi (2n+1)^{2} \tan^{-1} \left ( \frac{\pi \log(2)}{(2n+1)(2n) \pi^2 + \log(2)^2} \right ) - \\ \frac{\log(2)^2}{\pi} \tan^{-1} \left ( \frac{\pi \log(4)}{(2n+1)(2n-1)\pi^2+\log(2)^2}\right ) + \\ \log(2) \left \{ 2+(2n-1)\log[(2n-1)^2\pi^2 + \log(2)^2] + \\ 2\log[4n^2\pi^2+\log(2)^2]-(2n+1)\log[(2n+1)^2\pi^2+\log(2)^2]\right \} \right ] + K \tag{1} \label{eq1}$$

and

$$2 \operatorname{Li}_{2}\left(-\frac12 \right) + \frac{\pi^2}{6} = \sum_{n=1}^{\infty} \left[ \pi (2n-2)^2 \tan^{-1}\left( \frac{\pi \log(2)}{(2n-2)(2n-1)\pi^2+\log(2)^2}\right) + \\ 4\pi n^2\tan^{-1}\left( \frac{\pi \log(2)}{(2n-1)(2n)\pi^2+\log(2)^2} \right) - \\ \frac{\log(2)^2}{\pi}\tan^{-1}\left( \frac{\pi \log(4)}{(2n-2)(2n)\pi^2+\log(2)^2} \right) + \\ 2\log(2)\left\{ 1 + (n-1)\log[(2n-2)^2\pi^2+\log(2)^2] + \\ \log[(2n-1)^2\pi^2+\log(2)^2] -n \log[4n^2\pi^2+\log(2)^2] \right\} \right] \tag{2}\label{eq2}$$

In \eqref{eq1}, $$K = \pi \tan^{-1}\left( \frac{\pi}{\log(2)}\right) - \frac{\log(2)^2}{\pi} \tan^{-1}\left( \frac{\pi}{\log(2)}\right) + \log(2)[1+2\log(\log(2))-\log(\pi^2+\log(2)^2)] $$

We already know that $\operatorname{Li}_{2}\left(\frac12\right) = \frac{\pi^2}{12}-\frac{\log(2)^2}{2}$ and there is no known closed form for $\operatorname{Li}_{2}\left(-\frac12\right)$.

Obviously I have extracted these complex summations by reverse engineering.

Questions:

  1. If instead one were to start from the RHS summation terms and try to reach a closed form (not known apriori), what techniques or conversions or simplifications would be helpful in tackling that complex summation? (Not asking anyone to solve, since I already know the end result).
  2. Since the RHS of \eqref{eq1} and \eqref{eq2} have a very similar structure, I wonder if some summation results from the first (which has a closed form) can be somehow used to attempt a closed form for $\operatorname{Li}_{2}\left(-\frac12\right)$. I know this is a weak line of argument, since both the Dilogarithms look very similar from their basic definitions already (and one need not resort to these monstrosities). Thoughts? Feel free to skip this soft question if you just have some comments on my first question.
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  • $\begingroup$ There is a more simpler series for $\text{Li}_2\left(-\frac{1}{2}\right)$ since $\displaystyle \text{Li}_2\left(-\frac{1}{2}\right)=\frac{1}{2}\int_0^1\frac{\ln x}{1+\frac{x}{2}}dx$ $\endgroup$
    – FDP
    Commented Jul 5 at 3:54
  • $\begingroup$ What kind of closed-form are you expecting for $\text{Li}_2\left(-\frac{1}{2}\right)$? Empirically one knows this is not a rational combination of $\zeta(2)$ and $\ln 2$. $\endgroup$
    – FDP
    Commented Jul 5 at 3:58
  • $\begingroup$ Not expecting, but wishful thinking. Something like $\operatorname{Li}_{2}(\frac12)$ $\endgroup$
    – Srini
    Commented Jul 5 at 4:01
  • $\begingroup$ As i said, there is argument (not proof) that dilog(-1/2) is not equal to $a\ln 2+b\zeta(2)$ with $a,b$ rational numbers. $\endgroup$
    – FDP
    Commented Jul 5 at 4:04
  • $\begingroup$ $\operatorname{Li}_{2}(-\frac12)$ is equal to a rational combination of $\operatorname{Li}_{2}(\frac14)$, $\zeta(2)$,$\ln^2 2$. $\endgroup$
    – FDP
    Commented Jul 5 at 4:12

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