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Questions tagged [polylogarithm]

For questions about or related to polylogarithm functions.

49 votes
9 answers
3k views

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 ...
user153012's user avatar
  • 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}...
pisco's user avatar
  • 19.1k
43 votes
7 answers
2k views

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)}\...
Zaid Alyafeai's user avatar
42 votes
7 answers
7k views

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)\...
Lucian's user avatar
  • 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....
Vladimir Reshetnikov's user avatar
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 ...
Laila Podlesny's user avatar
36 votes
8 answers
2k views

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}^{...
mike's user avatar
  • 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(...
Ali Shadhar's user avatar
  • 25.8k
33 votes
4 answers
3k views

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^...
Oksana Gimmel's user avatar
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, ...
Kemono Chen's user avatar
  • 8,679
31 votes
3 answers
2k views

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 ...
Oksana Gimmel's user avatar
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}{...
pisco's user avatar
  • 19.1k
30 votes
3 answers
921 views

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{...
Shivam Patel's user avatar
  • 4,044
29 votes
4 answers
10k views

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-\...
X.C.'s user avatar
  • 2,602
29 votes
5 answers
1k views

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 ...
Nik Z.'s user avatar
  • 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(...
Anastasiya-Romanova 秀's user avatar
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 ...
David H's user avatar
  • 30.7k
26 votes
2 answers
1k views

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}...
user avatar
26 votes
2 answers
1k views

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}...
Vladimir Reshetnikov's user avatar
25 votes
4 answers
1k views

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(\...
Vladimir Reshetnikov's user avatar
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 ...
Renascence_5.'s user avatar
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}{\...
David H's user avatar
  • 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{...
user153012's user avatar
  • 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{...
Laila Podlesny's user avatar
23 votes
3 answers
2k views

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\{\...
Ali Shadhar's user avatar
  • 25.8k
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
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{{...
Shobhit Bhatnagar's user avatar
21 votes
1 answer
921 views

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\...
Vladimir Reshetnikov's user avatar
21 votes
1 answer
1k views

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 ...
Math Attack's user avatar
21 votes
3 answers
2k views

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 =...
pisco's user avatar
  • 19.1k

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