Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
546
questions
49
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9
answers
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Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $
I'm looking for a closed form of this integral.
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$
where $\operatorname{Li}_2$ is the dilogarithm function.
A numerical ...
- 12.4k
44
votes
2
answers
3k
views
Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$
We have the following result ($\text{Li}_{n}$ being the polylogarithm):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}...
- 19.1k
43
votes
7
answers
2k
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Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$
In the following thread
I arrived at the following result
$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$
Defining
$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\...
- 14.4k
42
votes
7
answers
7k
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Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$
$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?
Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\...
- 48.5k
41
votes
2
answers
2k
views
Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
Here is another integral I'm trying to evaluate:
$$I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx.\tag1$$
A numeric approximation is:
$$I\approx-0....
- 47.3k
37
votes
2
answers
4k
views
A difficult logarithmic integral ${\Large\int}_0^1\log(x)\,\log(2+x)\,\log(1+x)\,\log\left(1+x^{-1}\right)dx$
A friend of mine shared this problem with me. As he was told, this integral can be evaluated in a closed form (the result may involve polylogarithms). Despite all our efforts, so far we have not ...
- 13.2k
36
votes
8
answers
2k
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How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$
I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{...
- 667
35
votes
0
answers
2k
views
Are these generalizations known in the literature?
By using
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$
and
$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(...
- 25.8k
33
votes
4
answers
3k
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Integral ${\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx$
This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here:
Prove:
$$\int_0^1\ln(1-x)\ln(1+x)\ln^...
- 5,342
32
votes
1
answer
818
views
On the relationship between $\Re\operatorname{Li}_n(1+i)$ and $\operatorname{Li}_n(1/2)$ when $n\ge5$
Motivation
$\newcommand{Li}{\operatorname{Li}}$
It is already known that:
$$\Re\Li_2(1+i)=\frac{\pi^2}{16}$$
$$\Re\Li_3(1+i)=\frac{\pi^2\ln2}{32}+\frac{35}{64}\zeta(3)$$
And by this question, ...
- 8,679
31
votes
3
answers
2k
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What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$?
Some time ago I asked How to find $\displaystyle{\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$.
Thanks to great effort of several MSE users, we now know that
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln ...
- 5,342
30
votes
4
answers
2k
views
Conjectural closed-form of $\int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$
Let $$I_n = \int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$$
In a recently published article, $I_n$ are evaluated for $n\leq 6$:
$$\begin{aligned}I_1 &= \frac{\log ^2(2)}{2}-\frac{\pi ^2}{...
- 19.1k
30
votes
3
answers
921
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A closed form for a lot of integrals on the logarithm
One problem that has been bugging me all this summer is as follows:
a) Calculate
$$I_3=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \ln{(1-x)} \ln{(1-xy)} \ln{(1-xyz)} \,\mathrm{d}x\, \mathrm{d}y\, \mathrm{...
- 4,044
29
votes
4
answers
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Yet another log-sin integral $\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\sin x)\,dx$
There has been much interest to various log-trig integrals on this site (e.g. see [1][2][3][4][5][6][7][8][9]).
Here is another one I'm trying to solve:
$$\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\...
- 2,602
29
votes
5
answers
1k
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Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$
Please help me to evaluate this integral in a closed form:
$$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$
Using integration by parts I found that it could be expressed through ...
- 1,871
29
votes
7
answers
1k
views
Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$
I have spent my holiday on Sunday to crack several integral & series problems and I am having trouble to prove the following integral
\begin{equation}
\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(...
- 19.4k
28
votes
4
answers
1k
views
Integrals of integer powers of dilogarithm function
I'm interested in evaluating integrals of positive integer powers of the dilogarithm function. I'd like to see the general case tackled if possible, or barring that then as many particular cases as ...
- 30.7k
26
votes
2
answers
1k
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Show that $\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3){2n\choose n}}=\frac{9\pi^2}{8}$
I don't how prove this series and I have try look through maths world and Wikipedia on sum for help but no use at all, so please help me to prove this series.
How to show that
$$\sum_{n=0}^{\infty}...
user334593
26
votes
2
answers
1k
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Integral ${\large\int}_0^1\ln^3\!\left(1+x+x^2\right)dx$
I'm interested in this integral:
$$I=\int_0^1\ln^3\!\left(1+x+x^2\right)dx.\tag1$$
Can we prove that
$$\begin{align}I&\stackrel{\color{gray}?}=\frac32\ln^33-9\ln^23+36\ln3+2\pi^2\ln3-\frac{4\pi^2}...
- 47.3k
25
votes
4
answers
1k
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Integral involving Clausen function ${\large\int}_0^{2\pi}\operatorname{Cl}_2(x)^2\,x^p\,dx$
Consider the Clausen function $\operatorname{Cl}_2(x)$ that can be defined for $0<x<2\pi$ in several equivalent ways:
$$\begin{align}\operatorname{Cl}_2(x)&=-\int_0^x\ln\left(2\sin\left(\...
- 47.3k
25
votes
4
answers
1k
views
Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $
Evaluate
$$\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta $$
Several days ago,I found this interesting integral from a paper about generalized log-sine ...
- 5,281
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
23
votes
1
answer
533
views
Closed form for $\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}}$
Let
$$S=\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}},\tag1$$
where $\operatorname{Li}_a(z)$ is the polylogarithm. For $a=1/2$ it can be represented as
$$\begin{...
- 13.2k
23
votes
3
answers
2k
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Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$
The following problem is proposed by a friend:
$$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$
$$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\...
- 25.8k
22
votes
2
answers
3k
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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.
$$\operatorname{Re}\big[\operatorname{Li}_2\left(i\right)\big]=-\frac{\pi^2}{48},\hspace{1em}\...
- 1,928
22
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4
answers
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Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?
Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{...
- 11.6k
21
votes
1
answer
921
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Closed form for ${\large\int}_0^1\frac{\ln^3x}{\sqrt{x^2-x+1}}dx$
This is a follow-up to my earlier question Closed form for ${\large\int}_0^1\frac{\ln^2x}{\sqrt{1-x+x^2}}dx$.
Is there a closed form for this integral?
$$I=\int_0^1\frac{\ln^3x}{\sqrt{x^2-x+1}}dx\...
- 47.3k
21
votes
1
answer
1k
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Solution of a meme integral: $\int \frac{x \sin(x)}{1+\cos(x)^2}\mathrm{d}x$
Context
A few days ago I saw a meme published on a mathematics page in which they joked about the fact that $$\int\frac{x\sin(x)}{1+\cos(x)^2}\mathrm{d}x$$ was very long (and they put a screen shot of ...
21
votes
3
answers
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Evaluate $\int_0^1\arcsin^2(\frac{\sqrt{-x}}{2}) (\log^3 x) (\frac{8}{1+x}+\frac{1}{x}) \, dx$
Here is an interesting integral, which is equivalent to the title
$$\tag{1}\int_0^1 \log ^2\left(\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}\right) (\log ^3x) \left(\frac{8}{1+x}+\frac{1}{x}\right) \, dx =...
- 19.1k
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
$$\approx 111....
- 44.4k
20
votes
6
answers
775
views
A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Let $H_q$ denote harmonic numbers (generalized to a non-integer index $q$):
$$H_q=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+q}\right)=\int_0^1\frac{1-x^q}{1-x}dx=\gamma+\psi(q+1),\tag1$$
where $\psi(z)=\...
- 47.3k
20
votes
4
answers
1k
views
Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
- 5,626
20
votes
2
answers
1k
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Closed form for $\sum_{n=0}^\infty\frac{\Gamma\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}$
I was experimenting with hypergeometric-like series and discovered the following conjecture (so far confirmed by more than $5000$ decimal digits):
$$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\...
- 47.3k
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:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!...
- 47.3k
20
votes
1
answer
925
views
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
How to prove
$$\int_0^1x\ln^2(1+x)\ln\left(\frac{x^2}{1+x}\right)\frac{dx}{1+x^2}=-\frac{7}{32}\cdot\zeta{(3)}\ln2+\frac{3\pi^2}{128}\cdot\ln^22-\frac{1}{64}\cdot\ln^42-\frac{13\pi^4}{46080}$$
The ...
- 5,507
20
votes
1
answer
603
views
A conjectured identity for tetralogarithms $\operatorname{Li}_4$
I experimentally discovered (using PSLQ) the following conjectured tetralogarithm identity:
$$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
709
views
Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$
It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about
the cubic version, that is
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$
What are ...
- 44.4k
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
2
answers
1k
views
Conjectured closed form for $\operatorname{Li}_2\!\left(\sqrt{2-\sqrt3}\cdot e^{i\pi/12}\right)$
There are few known closed form for values of the dilogarithm at specific points. Sometimes only the real part or only the imaginary part of the value is known, or a relation between several different ...
- 47.3k
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
8
answers
1k
views
About the integral $\int_{0}^{1}\frac{\log(x)\log^2(1+x)}{x}\,dx$
I came across the following Integral and have been completely stumped by it.
$$\large\int_{0}^{1}\dfrac{\log(x)\log^2(1+x)}{x}dx$$
I'm extremely sorry, but the only thing I noticed was that the ...
17
votes
1
answer
1k
views
A rare integral involving $\operatorname{Li}_2$
A rare but interesting integral problem:
$$\int_{0}^{1}
\frac{\operatorname{Li}_2(-x)-
\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x)
-\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} }
=\...
- 5,370
17
votes
3
answers
825
views
A conjectured value for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$
In evaluating the integral given here it would seem that:
$$\operatorname{Re} \operatorname{Li}_4 (1 + i) \stackrel{?}{=} -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{...
- 11.8k
17
votes
2
answers
1k
views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
- 25.8k
17
votes
2
answers
834
views
Sum $\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$
I would like to seek your assistance in computing the sum
$$\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$$
I am stumped by this sum. I have tried summing the residues of $\displaystyle f(z)=\frac{\pi\...
- 5,604
17
votes
2
answers
894
views
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...
Question: How to show the relation
$$
J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt
=-\frac 1{64}\pi^4
$$
(using a "minimal industry" of relations, ...
- 34.2k
17
votes
1
answer
634
views
Polylogarithm ladders for the tribonacci and n-nacci constants
While reading about polylogarithms, I came across the nice polylogarithm ladder,
$$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
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 ...
- 14.4k
16
votes
3
answers
918
views
How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?
Can we evaluate $\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$ ?
where $H_n=\sum_{k=1}^n\frac1n$ is the harmonic number.
A related integral is $\displaystyle\int_0^1\frac{\ln^2(1-x)\...
- 25.8k