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10 votes
2 answers
509 views

Trilogarithm $\operatorname{Li}_3(z)$ and the imaginary golden ratio $i\,\phi$

I experimentally discovered the following conjectures: $$\Re\Big[1800\operatorname{Li}_3(i\,\phi)-24\operatorname{Li}_3\left(i\,\phi^5\right)\Big]\stackrel{\color{gray}?}=100\ln^3\phi-47\,\pi^2\ln\phi-...
Vladimir Reshetnikov's user avatar
5 votes
1 answer
753 views

On what domain is the dilogarithm analytic?

The series $\displaystyle\sum \dfrac{z^n}{n^2}$ converges for $\lvert z\rvert<1$ by the ratio test, meaning that the dilogarithm function $\text{Li}_2(z),$ which is equal to the series $\...
ziggurism's user avatar
  • 16.9k
4 votes
0 answers
301 views

Simplification of an expression involving the dilogarithm with complex argument

Do you think there is a way to get a nice form of the expression below $$\Im{\left( \text{Li}_2\left(\frac{3}{5}+\frac{4 i}{5}\right)- \text{Li}_2\left(-\frac{3}{5}+\frac{4 i}{5}\right)+ \text{Li}_2\...
user 1591719's user avatar
  • 44.4k
20 votes
3 answers
908 views

Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$

I numerically discovered the following conjecture: $$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!...
Vladimir Reshetnikov's user avatar
8 votes
2 answers
2k views

Find the derivative of a polylogarithm function

I was trying to find to which function the next series converges. $$ \sum_{n=1}^{\infty} \ln(n)z^n $$ If we take the polylogarithm function $Li_s(z)$ defined as $$ Li_s(s)=\sum_{n=1}^{\infty} \frac{z^...
Alonso Delfín's user avatar
4 votes
3 answers
170 views

There's a small detail in this proof on why $\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}$ that I can't figure out

http://www.maa.org/sites/default/files/pdf/upload_library/2/Kalman-2013.pdf Here is a link to the article I have been reading. It's really interesting and easy to follow. What bothers me is a result ...
Nick Freeman's user avatar
22 votes
2 answers
3k views

Extract real and imaginary parts of $\operatorname{Li}_2\left(i\left(2\pm\sqrt3\right)\right)$

We know that polylogarithms of complex argument sometimes have simple real and imaginary parts, e.g. $$\operatorname{Re}\big[\operatorname{Li}_2\left(i\right)\big]=-\frac{\pi^2}{48},\hspace{1em}\...
OlegK's user avatar
  • 1,928
4 votes
1 answer
347 views

On a property of polylogarithm

I have an observation, and I don't know that the following statement is true or not. If not give a counterexample, if it is true prove it, or give a reference about it. Let $n \in \mathbb{R}$, $z \in ...
user153012's user avatar
  • 12.4k
1 vote
1 answer
121 views

Inverse of Higher logarithms

Th polylogarithm function is defined by $$Li_s(z)=\sum_{k=1}^\infty\frac{z^k}{k^s}.$$ At $s=1$, we have the natural logarithm function. We have the inverse of natural logarithm function as the ...
Turbo's user avatar
  • 6,245
16 votes
5 answers
1k views

Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $

I proved the following result $$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$ After ...
Zaid Alyafeai's user avatar
15 votes
1 answer
228 views

Simplification of a trilogarithm of a complex argument

Is it possible to simplify the following expression? $$\large\Im\,\operatorname{Li}_3\left(-e^{\xi\,\left(\sqrt3-\sqrt{-1}\right)-\frac{\pi^2}{12\,\xi}\left(\sqrt3+\sqrt{-1}\right)}\right)$$ where $$\...
Marty Colos's user avatar
  • 3,320

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