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Is there a closed form expression for \begin{align} e^{\Large\frac{i\pi}3} \text{Li}_{2}\left( \frac{e^{\Large\frac{i\pi}3} }{2}\right) + e^{-\Large\frac{i\pi}3} \text{Li}_{2}\left( \frac{e^{-\Large\frac{i\pi}3} }{2}\right) \end{align} in terms of known constants such as: $\pi$, $\ln 2$, $\ln 3$, $e$, $\gamma$, etc.

The expression to be calculated can be placed into series form and is equivalent to calculating the series \begin{align} \sum_{n=1}^{\infty} \frac{ \cos\left( \frac{(n+1)\pi}{3} \right) }{2^{n-1} \ n^{2}} \end{align} in closed form.

Tunk-Fey
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Leucippus
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  • At least you can tell the result is a real number (plug the dilog series expansion in your expression). Also, there may be a clever Fourier series to use. – Jean-Claude Arbaut May 28 '14 at 06:20
  • @Jean-Claude I added to the problem for clarity of the infinite series form. It can be expressed in terms of the Lerch Zeta function, but in concept, that is trading one series for another. – Leucippus May 28 '14 at 06:29

1 Answers1

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Letting $z_{\pm}=\frac{e^{\pm\frac{i\pi}{3}}}{2}$, we find,

$$\frac{z_{\pm}}{z_{\pm}-1}=\mp\frac{i}{\sqrt{3}},$$

and

$$\ln{\left(1-z_{\pm}\right)}=\frac12\ln{\left(\frac34\right)}\mp i\frac{\pi}{6},\\ \implies \ln^2{\left(1-z_{\pm}\right)}=\frac14\ln^2{\left(\frac43\right)}-\frac{\pi^2}{36}\pm i \frac{\pi}{6}\ln{\left(\frac43\right)}.$$

Then using Landen's dilogarithm identity,

$$\begin{align} \operatorname{Li}_{2}{\left(z_{\pm}\right)} &=-\operatorname{Li}_{2}{\left(\frac{z_{\pm}}{z_{\pm}-1}\right)}-\frac12\ln^2{\left(1-z_{\pm}\right)}\\ &=-\operatorname{Li}_{2}{\left(\mp\frac{i}{\sqrt{3}}\right)}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)}\pm i \frac{\pi}{12}\ln{\left(\frac34\right)}. \end{align}$$

The real and imaginary components of $\operatorname{Li}_{2}{\left(z_{+}\right)}$ are then,

$$\Re{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}=-\Re{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)}\\ =-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)},$$

and,

$$\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}=-\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}+\frac{\pi}{12}\ln{\left(\frac34\right)}.$$

The term $\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}$ may be further simplified with the aid of the dilogarithmic identity,

$$\operatorname{Li}_{2}{\left(z\right)}-\operatorname{Li}_{2}{\left(-z\right)}+\operatorname{Li}_{2}{\left(\frac{1-z}{1+z}\right)}-\operatorname{Li}_{2}{\left(-\frac{1-z}{1+z}\right)}=\frac{\pi^2}{4}+\ln{\left(z\right)}\ln{\left(\frac{1+z}{1-z}\right)}.$$

Setting $z=-\frac{i}{\sqrt{3}}$,

$$\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}-\operatorname{Li}_{2}{\left(\frac{i}{\sqrt{3}}\right)}+\operatorname{Li}_{2}{\left(e^{\frac{i\pi}{3}}\right)}-\operatorname{Li}_{2}{\left(-e^{\frac{i\pi}{3}}\right)} = \frac{\pi^2}{12}+i\frac{\pi}{6}\ln{(3)}.$$

$$\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}-\operatorname{Li}_{2}{\left(\frac{i}{\sqrt{3}}\right)} = i\frac{\pi}{6}\ln{(3)}+i\frac{5\pi^2}{9\sqrt{3}}-i\frac{5}{6\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$

$$\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]} = \frac{\pi}{12}\ln{(3)}+\frac{5\pi^2}{18\sqrt{3}}-\frac{5}{12\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$

$$\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]} = -\frac{\pi}{6}\ln{(2)}-\frac{5\pi^2}{18\sqrt{3}}+\frac{5}{12\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$

The quantity of interest is then:

$$\begin{align} A &=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2z_{-}\operatorname{Li}_{2}{\left(z_{-}\right)}\\ &=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2\,\overline{z_{+}}\operatorname{Li}_{2}{\left(\overline{z_{+}}\right)}\\ &=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2\,\overline{z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}}\\ &=4\Re{\left[z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}\\ &=\Re{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}-\sqrt{3}\,\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}\\ &=-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{7\pi^2}{24}-\frac18\ln^2{\left(\frac34\right)}+\frac{\sqrt{3}\,\pi}{6}\ln{(2)}-\frac{5}{12}\psi^{(1)}{\left(\frac13\right)}.\\ \end{align}$$

I don't think the sum of dilogarithm trigamma terms in the final line can be simplified to a combination of the basic constants suggested.

David H
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