Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
0
votes
0answers
13 views
Patterns for the polylogarithm $\rm{Li}_m\big(\tfrac12\big)$ and Nielsen polylogarithm $S_{n,p}\big(\tfrac12\big)$?
The polylogarithm $\rm{Li}_m\big(\tfrac12\big)$ has closed-forms known for $m=1,2,3$. It seems this triad pattern extends to the Nielsen generalized logarithm $S_{n,p}\,\big(\tfrac12\big)$ for $n=0,1,...
4
votes
0answers
93 views
More on the log sine integral $\int_0^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta$
I. In this post, the OP asks about the particular log sine integral,
$$\mathrm{Ls}_{7}^{\left ( 3 \right )} =-\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\...
1
vote
0answers
79 views
Closed-forms for the integral $\int_0^1\frac{\rm{Li}_n(x)}{1+x}dx$?
(This is related to this question.)
Define the integral,
$$I_n = \int_0^1\frac{\rm{Li}_n(x)}{1+x}dx$$
with polylogarithm $\rm{Li}_n(x)$. Given the Nielsen generalized polylogarithm $S_{n,p}(z)$,
$$...
5
votes
0answers
140 views
More on the integral $\int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$
In this post, the OP asks about the integral,
$$I = \int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$$
I. User DavidH gave a beautiful (albeit long)...
2
votes
1answer
93 views
Integral: $\int_0^1\frac{\mathrm{Li}_2(x^2)}{\sqrt{1-x^2}}dx$
I am trying to evaluate $$P=\frac\pi2\sum_{n\geq1}\frac{{2n\choose n}}{4^n n^2}$$
I used the beta function to show that
$$P=\int_0^1\frac{\mathrm{Li}_2(x^2)}{\sqrt{1-x^2}}dx$$
IBP:
$$P=\sin^{-1}(x)\...
4
votes
0answers
142 views
A twisted hypergeometric series $\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2$
I was given that $$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\frac{(2n)!}{4^n(n!)^2}\right)^2=\frac{32}\pi G\ln2+\frac{64}\pi\Im\operatorname{Li}_3\left(\frac{1+i}2\right)-2\ln^22-\frac53\pi^2,$$ where $...
3
votes
4answers
119 views
How can I compute this integral in closed form : $\int_0^{\frac{π}{4}}\ln^2(\tan x)dx$
How can I compute this integral in closed form :
$\displaystyle\int_0^{\displaystyle \tfrac{π}{4}}\ln^2(\tan x)dx$
How can use Fourier series here ?
$-2\displaystyle \sum_{n=0}^{\infty}\frac{\cos(...
-1
votes
2answers
75 views
Compute in closed form : $\int_0^{\frac{π}{4}} x\ln(\tan x)\left(1-\frac{1}{\cos^2 x}\right)dx$
Question :
Compute in closed form without use series
$I =\displaystyle\int_0^{\pi / 4} x\ln\left(\tan x\right)\left(1-\frac{1}{\cos^2 x}\right)\,dx$
I think use : $y=\tan x$ then $dy=\frac{1}{\cos^...
2
votes
3answers
118 views
Compute in closed form $\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$
I am trying to find closed form for this integral:
$$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$
Where $a>0$.
My try: Let: $$I(a)=\int_0^1\frac{\arctan{ax}}{\sqrt{1-x^{2}}}dx$$
Then:
$$\...
5
votes
2answers
118 views
Compute this following integral without Fourier series : $\int_0^{\pi/4}x\ln(\tan x)dx$
Compute the following integration without harmonic series or Fourier series :
$I=\displaystyle\int_0^{\frac{π}{4}}x\ln(\tan x)dx$
Wolfram alpha give $I=\frac{7\zeta(3)-4πC}{16}$
Where $C$ : Catalan'...
2
votes
0answers
93 views
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?
I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
0
votes
0answers
33 views
Closed form for a family of definite integrals involving a Gaussian and error functions.
Let $n\ge 0$ be an integer and let $c \in {\mathbb R}$.
Let us define:
\begin{eqnarray}
{\mathfrak F}^{(A,B)}_{a,b} &:=& \int\limits_A^B \frac{\log(z+a)}{z+b} dz\\
&=& F[B,a,b] - F[A,...
0
votes
0answers
28 views
An anti-derivative involving an arc tangent, a square root and a rational function.
This question is similar to A generalized Ahmed's integral .
Let $a_1 \in {\mathbb R}_+$, $a_2 \in {\mathbb R}_+$ and $b_1 \in {\mathbb R}_+$. Consider the following integrals:
\begin{eqnarray}
{\...
6
votes
3answers
142 views
On $\int_{-\pi/2}^{\pi/2}\operatorname{Li}_3(\sin x)dx$ and its derivative
Question 1: How can we prove $$I_3=\int_{-\pi/2}^{\pi/2}\operatorname{Li}_3(\sin x)dx=\frac{\pi}{4}\zeta(3)+\frac{1}{6}\pi\ln^32-\frac{1}{24}\pi^3\ln2?$$
(where $\displaystyle\operatorname{Li}_s(x):=...
13
votes
1answer
279 views
Yet another difficult logarithmic integral
This question is a follow-up to MSE#3142989.
Two seemingly innocent hypergeometric series ($\phantom{}_3 F_2$)
$$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{(-1)^n}{2n+1}\qquad \...
13
votes
1answer
300 views
A logarithmic integral, generalization of a result of Shalev
As many of you are already aware, I and Marco Cantarini are currently working on the applications of fractional operators to hypergeometric series, extending the class of $\phantom{}_{p+1} F_p$s whose ...
1
vote
1answer
78 views
On $\sum_{k=1}^\infty1/(k!k^s)$.
Has the following $\zeta$-like function been studied before?
$$f(z;s)=\sum_{k=1}^\infty\frac{z^k}{k!k^s}.$$
I believe this is an entire function since using the ratio test,
$$\lim_{n\to\infty}\frac{...
9
votes
1answer
203 views
A peculiar Euler sum
I would like a hand in the computation of the following Euler sum (Why isn't here a tag for Euler sums?)
$$ S=\sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2m+1)(2n+1)^2(2m+2n+1)} \tag{1}$$
which arises from ...
3
votes
0answers
88 views
Evaluating $\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_n$
I am trying to evaluate the following integral
$$\int_{0}^{1-x} \frac{1}{1-x_{n}} \cdots \int_{0}^{1-x_3} \frac{1}{1-x_2} \int_{0}^{1-x_2} \frac{1}{1-x_1} dx_1 dx_2 \cdots dx_{n}, \hspace{0.5cm} 0<...
3
votes
0answers
43 views
Can every Gaussian integral be reduced to elementary functions and poly-logarithms only?
Let us define a following function:
\begin{eqnarray}
{\mathcal J}^{(d)}(\vec{A}) := \int\limits_0^\infty e^{-u^2} \prod\limits_{\xi=1}^d erf(A_\xi u) du
\end{eqnarray}
for $\vec{A}:=\left(A_\xi\right)...
5
votes
0answers
99 views
Is a closed form possible for $\int\frac{\text{Li}_2(x)^2}{x}dx$?
Can $\,\displaystyle\int\frac{\text{Li}_2(x)^2}{x}dx\,$ be calculated
by a sum/term of polylogarithm functions and the natural logarithm and polynomials (“closed form”) ?
For the special case $\,\...
0
votes
1answer
91 views
An integral involving a Gaussian, error functions and the Owen's T function.
This question is closely related to An integral involving a Gaussian and an Owen's T function. and An integral involving error functions and a Gaussian .
Let $\nu_1 \ge 1$ and $\nu_2 \ge 1$ be ...
5
votes
1answer
90 views
Polylogarithm inequality: $(s+1)\frac{-\operatorname{Li}_{s+1} (-x)}{-\operatorname{Li}_s(-x)} > \log(x)$
For $s \geq 0$ and $x > 0$ define
$$ f_s (x) = - \operatorname{Li}_s (-x) \stackrel{s > 0}{=} \frac{1}{\Gamma(s)} \int \limits_0^\infty \frac{x t^{s-1}}{\mathrm{e}^t + x} \, \mathrm{d} t \, .$$
...
4
votes
1answer
53 views
Series power function over exponential function
A typical exercise from calculus is to show that any exponential function eventually grows faster than any power function, i.e.
$$ \lim_{k \to \infty} \frac{k^a}{b^k} = 0 \qquad \text{ for } a,b>1.$...
4
votes
2answers
124 views
I need help computing $\int {\ln x\over 2-x}\, dx$
While integrating $\ln(\sec x)$, at one point I managed to break the integral into two. But I wasn't able to integrate one of those parts.
The integral I am having a difficulty with is:
$$\int {\ln ...
7
votes
1answer
144 views
closed form for $\int_0^1\frac{\mathrm{Li}_s(x-x^2)}{x-x^2}\mathrm dx$
I am trying to evaluate
$$F(s)=\sum_{n\geq1}\frac1{n^{s+1}{2n\choose n}}$$
I started off by noting that $$\frac1{n{2n\choose n}}=\frac12\int_0^1\left[x-x^2\right]^{n-1}\mathrm dx$$
So
$$F(s)=\int_0^1\...
0
votes
1answer
40 views
Polylogarithm grows slower than polynomial proof
In the CLRS book, there's this part, where it's shown that $$\lim_{n\to\infty}\frac{(n^b)}{(a^n)} = 0$$. In the same chapter, it uses the aforementioned equation to prove that any polylogarithm ...
12
votes
6answers
374 views
Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)dx=\frac43G+\frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\...
4
votes
3answers
74 views
Prove $\text{Li}_2(e^{-2 i x})+\text{Li}_2(e^{2 i x})=\frac{1}{3} (6 x^2-6 \pi x+\pi ^2)$ when $0<x<\pi$
This is an identity I deduced when playing with the initial-boundary value problem of heat conduction equation asked here. It's easy to verify numerically with Mathematica:
...
2
votes
0answers
65 views
Polylogarithmic integrals
I'm a physicist looking at the Fredholm inverse of some integral equation. In attempting to solve the equation I stumbled upon a type of integral of the form
\begin{equation}
\int \frac{\prod_{i=1}^N \...
3
votes
1answer
41 views
Polylog property $\operatorname{Li}_n(1-i)+(-1)^n\operatorname{Li}_n(\frac{1+i}2)$
How can we calculate $$L_n=\operatorname{Li}_n(1-i)+(-1)^n\operatorname{Li}_n\left(\frac{1+i}2\right)?$$
I have found the following result manually:
$$L_1=-\frac12\pi i+\ln\frac{1-i}2=-\frac34\pi i-\...
2
votes
1answer
64 views
Logarithmic integrals and Euler sums
At various places e.g.
Calculate $\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$
and
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
logarithmic integrals are connected ...
12
votes
0answers
328 views
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
How to prove$$\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}$$
The ...
1
vote
0answers
66 views
Upper bound the Polylogarithm $\sum_{n=1}^\infty \frac{x^n}{n^2}$
Let $x \in (0,1)$ be some real number, we can then consider the Polylogarithm:
$$\operatorname{L}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$$
It is not hard to see that the following upper bound holds:
$$...
10
votes
1answer
429 views
Evaluate$\int\limits_0^1 [\log(x)\log(1-x)+\operatorname{Li}_2(x)]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx$
$$\mathfrak{I}=\int\limits_0^1 \left[\log(x)\log(1-x)+\operatorname{Li}_2(x)\right]\left[\frac{\operatorname{Li}_2(x)}{x(1-x)}-\frac{\zeta(2)}{1-x}\right]\mathrm dx=4\zeta(2)\zeta(3)-9\zeta(5)\tag1$$
...
4
votes
1answer
151 views
Proving $\Im\operatorname{Li}_2(\sqrt i(\sqrt 2-1))=\frac34G+\frac18\pi\ln(\sqrt2-1)$
$\newcommand{\Li}{\operatorname{Li}_2}$
I found, numerically, that $$\Im\Li(\sqrt i(\sqrt 2-1))=\frac34G+\frac18\pi\ln(\sqrt2-1).$$
How can we prove it?
My attempt of proving this equation:
...
3
votes
0answers
57 views
Are there any non-trivial special values of $\operatorname{Li}_4(z)$?
Denote $\operatorname{Li}_4(z)$ the analytic continuation of $\sum_{n=1}^\infty\frac{z^n}{n^4}$. $z$ is a algebraic number with $|z|\ne 0,1$. Does $\Re\operatorname{Li}_4(z)$ or $\Im\operatorname{Li}...
3
votes
4answers
196 views
Closed form of $\int{\lfloor{x}\rfloor}dx$
I calculated $\int{\lfloor{x}\rfloor}dx$ and i got this result:
$$\int{\lfloor{x}\rfloor}dx = \frac{x^2-x}{2}+\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2+c$$
Do you know if this series ...
1
vote
1answer
69 views
What is the value of this expression involving the dilogarithm $\operatorname{Li}_2$ and $\sqrt{2}$?
I'm carrying out a calculation, and the end result is $$\operatorname{Li}_2(1/\sqrt2) - \operatorname{Li}_2(1 - 1/\sqrt2) +
\operatorname{Li}_2(2 - \sqrt2) - \operatorname{Li}_2(\sqrt{...
1
vote
0answers
51 views
Integrate $\int_{-\infty}^\infty [4(\log r_1 - \log r_2) - 2(x_1^2/r_1^2 - x_2^2/r_2^2)]^2 dx$
As the title suggests, I am having trouble evaluating the following definite integral:
$$\int_{-\infty}^\infty \left[4\left(\log r_1 - \log r_2\right) - 2\left(\frac{x_1^2}{r_1^2} - \frac{x_2^2}{r_2^...
-2
votes
2answers
147 views
Improper integral $\int_0^\infty \frac{x^{\alpha}\ln x}{x^2+1}\,dx=\frac{\pi^2}{4} \frac{\sin(\pi \alpha/2)}{\cos^2(\pi \alpha/2)}$ [closed]
$$\int_0^\infty \frac{x^{\alpha}\ln x}{x^2+1}\,dx=\frac{\pi^2}{4} \frac{\sin(\pi \alpha/2)}{\cos^2(\pi \alpha/2)}$$
where $0 < \alpha < 1$.
Answer: When i put this term in my integral ...
8
votes
2answers
113 views
How evaluate $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$
How prove
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$
$\mathbf {My\,Attempt:}$
...
1
vote
0answers
100 views
Evaluating $\int_0^1\frac{\ln(x)}{1-x/2}\int_0^{x/2}\frac{\ln(1-t)}{t}\,dt\,dx$
According to this post
$$Q=\int_0^1\frac{\ln(x)}{1-x/2}\int_0^{x/2}\frac{\ln(1-t)}{t}\,dt\,dx=\frac{1}{8}\zeta \left( 4 \right) - \frac{1}{2}\zeta \left( 2 \right){\ln ^2}(2) + \frac{1}{{12}}{\ln ^4}(...
1
vote
1answer
56 views
Evaluating $\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n^2}$ [duplicate]
According to Wolfram Alpha
$$\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n^2} = \frac{\pi^2}{12}-\frac{\ln^22}{2}$$
I searched on Wikipedia and learnt that
$$\sum_{n=1}^{\infty} \frac{\left(...
1
vote
0answers
40 views
Real Part of the Dilogarithm
It is well known that
$$\frac{x-\pi}{2}=-\sum_{k\geq 1}\frac{\sin{kx}}{k}\forall x\in(0,\tau),$$
which gives
$$\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}=\sum_{k\geq 1}\frac{\cos(kx)}{k^2}.$$
...
5
votes
1answer
132 views
On the series expansion of $\frac{\operatorname{Li}_3(-x)}{1+x}$ and its usage
Recently I have asked about the evaluation of an integral involving a Trilogarithm $($which can be found here$)$. Pisco provided a quite elegant approach starting with a functional equation of the ...
1
vote
2answers
60 views
Asymptotic expansion of $\operatorname{Li}_{-x}(1/2)$
I want to find the asymptotic expansion of $$\operatorname{Li}_{-x}(1/2):=\sum_{n=1}^\infty\frac{n^x}{2^n}$$as $\Re x\to+\infty$.
My Attempt
Firstly, I considered series $$f(y)=\sum_{m=1}^\infty \...
11
votes
2answers
281 views
Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$
Recently I have encountered some integrals involving Polylogarithms like this one or closely related integrals such as this one. Hence I am quite fascinated by these kinds of definite integrals $-$ ...
6
votes
2answers
223 views
Show that $\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}$
Recently I have come across the following integral while going over this list (Problem $35$)
$$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(...
2
votes
2answers
87 views
Dilogarithm property
I'm working on dilogarithms (Don Zagier 'The dilogarithm function', Matilde Lalìn 'dilogarithm, a cool function' )and I have encountered this six term relation:
$Li_2(x)+Li_2(y)+Li_2(z)=(1/2)[Li_2(-...
