2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

1
$\begingroup$

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

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

$\endgroup$
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
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.