# Simplify the difference of two dilogarithms--as in the logarithmic counterpart

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 $$\mbox{Li}_{n}$$.)

• 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 $\text{Li}_2\left(\frac{1}{18} \left(9+\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right)-\text{Li}_2\left(\frac{1}{18} \left(9-\sqrt{81-\frac{64}{\sqrt{3}}}\right)\right) \approx 1.08747$. Dec 6, 2019 at 18:59
• this was a typo, which I have corrected in the answer box. Dec 6, 2019 at 19:02

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

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

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

Mathematica's FullSimplify[] command can do nothing with the expression in question: So, it appears unlikely that this expression can be simplified.

• 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. Dec 6, 2019 at 16:00
• 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. Dec 6, 2019 at 16:23
• @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 $(\ln^2 t)/2+C$. Dec 6, 2019 at 18:30
• thanks, @IosifPinelis --- corrected in the answer box (too bad comments cannot be edited) Dec 6, 2019 at 19:05