Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
546
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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
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2
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The most complete reference for identities and special values for polylogarithm and polygamma functions
I am looking for a book, paper, web site, etc. (or several ones) containing the most complete list of identities and special values for the polylogarithm $\operatorname{Li}_s(z)$ and polygamma $\psi^{(...
- 47.3k
14
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5
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Integral $\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$
I would like to know how to evaluate the integral
$$\int^1_0\frac{\ln{x} \ \mathrm{Li}_2(x)}{1-x}dx$$
I tried expanding the integrand as a series but made little progress as I do not know how to ...
- 5,604
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Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$
Prove that
I encountered this integral while working on the sum $\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$. Both of the integral and the sum were proposed by Cornel Valean:
The ...
- 25.8k
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2
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How to evaluate $\int_0^\infty\frac{\frac{\pi^2}{6}-\operatorname{Li}_2\left(e^{-x}\right)-\operatorname{Li}_2\left(e^{-\frac{1}{x}}\right)}{x}dx$
I need to evaluate the following integral with a high precision:
$$
I=\int_{0}^{\infty}\left[%
{\pi^{2} \over 6} - {\rm Li}_2\left({\rm e}^{-x}\right)
-{\rm Li}_2\left({\rm e}^{-1/x}\right)\right]\,{{\...
- 1,871
14
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2
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Definite Dilogarithm integral $\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx $
Prove the following
$$\int^1_0 \frac{\operatorname{Li}_2^2(x)}{x}\, dx = -3\zeta(5)+\pi^2 \frac{\zeta(3)}{3}$$
where
$$\operatorname{Li}^2_2(x) =\left(\int^x_0 \frac{\log(1-t)}{t}\,dt \right)^2$$
- 14.4k
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Inverse of the polylogarithm
The polylogarithm can be defined using the power series
$$
\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s}.
$$
Contiguous polylogs have the ladder operators
$$
\operatorname{Li}_{s+1}(z) ...
- 1,136
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2
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Closed- form of $\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$
Is there a possibility to find a closed-form for
$$\int_0^1 \frac{{\text{Li}}_3^2(-x)}{x^2}\,dx$$
- 5,507
14
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1
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Closed-form of $\int_{0}^{\infty} \frac{{\text{Li}}_2^3(-x)}{x^3}\,dx$
Is there a possibility to find a closed-form for $$\int_{0}^{\infty} \frac{{\text{Li}}_2^3(-x)}{x^3}\,dx$$
We have
$$I=\int_0^1\frac{Li_2^3(-x)+x^4Li_2^3(-\frac{1}{x})}{x^3}\,dx$$
After repeatedly ...
- 5,507
14
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1
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Closed form for $\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}$
(This is a slight variation of another question, already answered)
Can we find a closed form of the following series?
$$S=\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}\tag1$$
Using some non-rigorous ...
- 47.3k
14
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1
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A peculiar Euler sum
I would like a hand in the computation of the following Euler sum
$$ S=\sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2m+1)(2n+1)^2(2m+2n+1)} \tag{1}$$
which arises from the computation of $\int_{0}^{1}\frac{\...
- 355k
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Yet another difficult logarithmic integral
This question is a follow-up to MSE#3142989.
Two seemingly innocent hypergeometric series ($\phantom{}_3 F_2$)
$$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{(-1)^n}{2n+1}\qquad \...
- 355k
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6
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Strategies for evaluating sums $\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n}$
I'm looking for strategies for evaluating the following sums for given $z$ and $m$:
$$
\mathcal{S}_m(z):=\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n},
$$
where $H_n^{(m)}$ is the generalized harmonic ...
- 12.4k
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3
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Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$
While going through the recent questions concerning tagged polylogarithms I stumbled upon this post which asks for a concrete evaluating of a polylogarithmic integral. However the post also states the ...
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Ways to prove $ \int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$?
I am wondering if we can show in a simple way that
$$
I=\int_0^1 \frac{\ln^2(1+x)}{x}dx = \int_1^2 \frac{\ln^2(t)}{t-1}dt = \frac{\zeta(3)}{4}.
$$
Because the end result is very simple, I suspect ...
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