This question--pertaining to the quantum-information-theoretic topic of "bound entanglement"--stems from the question and answer to https://math.stackexchange.com/questions/3464105/if-possible-meaningfully-simplify-an-expression-involving-logs-polylogs-and-hy/3465129#3465129

The answer (that is, the formula for the "bound entanglement probability") there contains the expression \begin{equation} \text{Li}_2\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_2\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right), \end{equation} where the polylogarithm (in particular, dilogarithm) is indicated.

We have further observed as part of the simplification analysis in that answer--changing the subscript of Li from 2 to 1 (leading to the standard logarithmic framework)--that \begin{equation} \text{Li}_1\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_1\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right) = \end{equation} \begin{equation} \log \left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\log \left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)= \end{equation} \begin{equation} 2 \coth ^{-1}\left(\frac{9}{\sqrt{81-\frac{64}{\sqrt{3}}}}\right), \end{equation}

So, is a "parallel" simplification possible in the original $\mbox{Li}_{2}$ case? (Of course, one might ask--more generally--about $\mbox{Li}_{n}$.)

  • $\begingroup$ I did the indicated integration and got $\frac{1}{2} \left(\log ^2\left(\frac{54}{27+\sqrt{729-192 \sqrt{3}}}\right)-\log ^2\left(-\frac{54}{\sqrt{729-192 \sqrt{3}}-27}\right)\right) \approx -2.05143$, while $\text{Li}_2\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_2\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right) \approx 1.08747$. $\endgroup$ Dec 6, 2019 at 18:59
  • $\begingroup$ this was a typo, which I have corrected in the answer box. $\endgroup$ Dec 6, 2019 at 19:02

2 Answers 2


$$\text{Li}_2\left(\tfrac{1}{2}+a\right)-\text{Li}_2\left(\tfrac{1}{2}-a\right)=-\int_{1/2-a}^{1/2+a}\frac{\log(1-t)}{t}\,dt,\;\;0<|{\rm Re}\,a|<1/2.$$

the integral of $\log(1-t)/t$ cannot be expressed in terms of elementary functions; it can be written in terms of special functions, but that brings us back to the polylog expression in the OP.

more generally, for any integer $n\in\mathbb{Z}$, $$f_n(a)=\text{Li}_n\left(\tfrac{1}{2}+a\right)-\text{Li}_n\left(\tfrac{1}{2}-a\right)=\int_{1/2-a}^{1/2+a}\frac{{\rm Li}_{n-1}(t)}{t}\,dt,\;\;0<|{\rm Re}\,a|<1/2,$$ and since ${\rm Li}_{n-1}(t)$ is a rational function for $n\leq 1$ this gives a simplification of $f_n(a)$ in terms of elementary functions when $n=1,0,-1,-2,\ldots$.

For example, $$f_1(a)=2 \tanh ^{-1}(2 a),\;\;f_0(a)=\frac{8 a}{1-4 a^2},$$ $$f_{-1}(a)=\frac{8 a \left(4 a^2+3\right)}{\left(1-4 a^2\right)^2},\;\;f_{-2}(a)=\frac{8 a \left(16 a^4+72 a^2+13\right)}{\left(1-4 a^2\right)^3}.$$

  • $\begingroup$ OK--checks out now! $\endgroup$ Dec 6, 2019 at 19:07
  • $\begingroup$ So, "philosophically" speaking, why the rather amazing hyperbolic cotangent simplification/identity in the (mono)logarithmic case? Perhaps, unanswerable. $\endgroup$ Dec 6, 2019 at 19:13
  • $\begingroup$ I have added the explanation why $n=1$ simplifies in terms of elementary functions. $\endgroup$ Dec 6, 2019 at 19:30

Mathematica's FullSimplify[] command can do nothing with the expression in question:

enter image description here

So, it appears unlikely that this expression can be simplified.

  • $\begingroup$ Thanks! Had tried this Mathematica command. Playing around with creating a series $\frac{\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)^k-\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)^k}{k^2}$ of the terms in the infinite series definition of the dilogarithm, then using the FindSequenceFunction command (a favorite "toy" of mine). I get DifferenceRoot results--but nothing yet of real use. $\endgroup$ Dec 6, 2019 at 16:00
  • 1
    $\begingroup$ since you are basically asking for an integral of $\log t/t$, an answer in terms of elementary functions is unlikely; and an answer in terms of special functions is what you have. $\endgroup$ Dec 6, 2019 at 16:23
  • 1
    $\begingroup$ @CarloBeenakker : You probably meant an integral of $\text{Li}_1(t)/t$, rather than a (say indefinite) integral of $\ln t/t$ -- which latter is $(\ln^2 t)/2+C$. $\endgroup$ Dec 6, 2019 at 18:30
  • $\begingroup$ thanks, @IosifPinelis --- corrected in the answer box (too bad comments cannot be edited) $\endgroup$ Dec 6, 2019 at 19:05

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