All Questions
Tagged with polylogarithm special-functions
119
questions
25
votes
2
answers
729
views
Definite integral of arcsine over square-root of quadratic
For $a,b\in\mathbb{R}\land0<a\le1\land0\le b$, define $\mathcal{I}{\left(a,b\right)}$ by the integral
$$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{a}\frac{\arcsin{\left(2x-1\right)}\,\mathrm{d}x}{\...
- 30.7k
24
votes
2
answers
927
views
Closed-form of $\int_0^1 \operatorname{Li}_3^3(x)\,dx$ and $\int_0^1 \operatorname{Li}_3^4(x)\,dx$
We know a closed-form of the first two powers of the integral of trilogarithm function between $0$ and $1$. From the result here we know that
$$I_1=\int_0^1 \operatorname{Li}_3(x)\,dx = \zeta(3)-\frac{...
- 12.4k
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.
$$\operatorname{Re}\big[\operatorname{Li}_2\left(i\right)\big]=-\frac{\pi^2}{48},\hspace{1em}\...
- 1,928
22
votes
4
answers
1k
views
Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?
Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{...
- 11.6k
20
votes
7
answers
905
views
Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$
Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$
Numerically, it's about
$$\approx 111....
- 44.4k
20
votes
1
answer
603
views
A conjectured identity for tetralogarithms $\operatorname{Li}_4$
I experimentally discovered (using PSLQ) the following conjectured tetralogarithm identity:
$$720 \,\text{Li}_4\!\left(\tfrac{1}{2}\right)-2160 \,\text{Li}_4\!\left(\tfrac{1}{3}\right)+2160 \,\text{Li}...
- 47.3k
19
votes
3
answers
948
views
Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$
I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed ...
- 11.6k
18
votes
3
answers
930
views
Closed form for $\int_0^e\mathrm{Li}_2(\ln{x})\,dx$?
Inspired by this question and this answer, I decided to investigate the family of integrals
$$I(k)=\int_0^e\mathrm{Li}_k(\ln{x})\,dx,\tag{1}$$
where $\mathrm{Li}_k(z)$ represents the polylogarithm of ...
- 856
17
votes
1
answer
634
views
Polylogarithm ladders for the tribonacci and n-nacci constants
While reading about polylogarithms, I came across the nice polylogarithm ladder,
$$6\operatorname{Li}_2(x^{-1})-3\operatorname{Li}_2(x^{-2})-4\operatorname{Li}_2(x^{-3})+\operatorname{Li}_2(x^{-6}) = ...
- 54.9k
16
votes
1
answer
641
views
The log integrals $\int_{0}^{1/2} \frac{\log(1+2x) \log(x)}{1+x} \, dx $ and $ \int_{0}^{1/2} \frac{\log(1+2x) \log(1-x)}{1+x} \, dx$
In attempting to evaluate $ \int_{0}^{\infty} [\text{Ei}(-x)]^{4} \, dx$ (which can be evaluated in terms of polylogarithm values), I determined that $$ \begin{align} \int_{0}^{\infty} [\text{Ei}(-x)]...
- 42.7k
16
votes
1
answer
551
views
Relations connecting values of the polylogarithm $\operatorname{Li}_n$ at rational points
The polylogarithm is defined by the series
$$\operatorname{Li}_n(x)=\sum_{k=1}^\infty\frac{x^k}{k^n}.$$
There are relations connecting values of the polylogarithm at certain rational points in the ...
- 47.3k
15
votes
3
answers
2k
views
Simplification of an expression containing $\operatorname{Li}_3(x)$ terms
In my computations I ended up with this result:
$$\mathcal{K}=78\operatorname{Li}_3\left(\frac13\right)+15\operatorname{Li}_3\left(\frac23\right)-64\operatorname{Li}_3\left(\frac15\right)-102
\...
- 5,342
15
votes
2
answers
928
views
Compute polylog of order $3$ at $\frac{1}{2}$
How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{1}{2^nn^3}$$
I am aware this equals polylog of order $3$ at $\frac{1}{2}$ or $\operatorname{Li}_3\left(\frac{1}{2}\right)$, but how ...
- 11k
15
votes
1
answer
228
views
Simplification of a trilogarithm of a complex argument
Is it possible to simplify the following expression?
$$\large\Im\,\operatorname{Li}_3\left(-e^{\xi\,\left(\sqrt3-\sqrt{-1}\right)-\frac{\pi^2}{12\,\xi}\left(\sqrt3+\sqrt{-1}\right)}\right)$$
where
$$\...
- 3,320
15
votes
1
answer
767
views
Known exact values of the $\operatorname{Li}_3$ function
We know some exact values of the trilogarithm $\operatorname{Li}_3$ function.
Known real analytic values for $\operatorname{Li}_3$:
$\operatorname{Li}_3(-1)=-\frac{3}{4} \zeta(3)$
$\operatorname{Li}...
- 12.4k