Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
416
questions
6
votes
2answers
223 views
Derive $\int_0^1 \frac{\ln(\sqrt2-1)-(\sqrt2-x)\ln x}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1) $
I obtained the integral
$$\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1)
$$
as a by-product while carrying out some complex analysis on an ...
0
votes
0answers
9 views
Clarification on analytic continuation of polylogarithm definition
I am trying to understand the branching geometry of the Dilogarithm function. In the Google book:Functional equation of hyperlogarithms, equation 8.6 written for the special case $n=2$ gives:
$$
\...
0
votes
3answers
98 views
Stuck with integral involving Polylogarithms $\int_{0}^{\infty}\frac{2t}{e^{t\pi}+1} \,dt$
For a research work I ended up needing to give a proof of Zeta's trivial zeros, in order to do so I tried using the Abel-Plana formula.
$\zeta(s)=\frac{2^{s-1}}{s-1}-2^{s}\int_{0}^{\infty} \frac{\sin(...
3
votes
1answer
164 views
Evaluate $\int^1_0 x^a (1-x)^b \operatorname{Li}_2 (x)\, \mathrm dx$
For what $a,b$ the integral
$$\int^1_0 x^a(1-x)^b\operatorname{Li}_2 (x)\, \mathrm dx$$
has a closed form solution? I tried to solve it by expanding dilogarithm function, or by reducing it to linear ...
0
votes
0answers
50 views
Using Contour Integral to find the value of $\int_{-1}^{+1}\frac{\ln{(1+t)}}{t}dt$
$\newcommand{LogI}{\operatorname{Li}}$
We know that the value of $\LogI_{2}(-1)$ is -$\frac{\pi^2}{12}$ and $\LogI_{2}(1)$ is $\frac{\pi^2}{6}$. The value of the polylogarithms has already been ...
1
vote
1answer
90 views
Definite integral for $\zeta(3)$
By making use of Mathematica, I detected the following integral expression for zeta(3):
$$\int_0^1\frac{\log(x)}{1+x}\log\left(\frac{2+x}{1+x}\right)dx=-\frac5{12}\zeta(3).$$
Any proof of it would be ...
0
votes
0answers
34 views
Analytical continuation for polylogarithm
It is known that the series presentation
$$L_s(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^s}$$
for the polylogarithm is valid only in the open disk $|z|<1$. Outside this region, the polylogarithm is ...
0
votes
0answers
27 views
Integral of a modified softplus function
In a manuscript I am currently reading, the authors propose a modified softplus function
$$g(a)=\frac{\log\left(2^a +1 \right)}{\log(2)}$$
for some $a \in \mathbb{R}$. The authors then claim that if $...
4
votes
0answers
261 views
How to evaluate $\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$
I am trying to evaluate
$$\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$$
But I am not sure what to do since integration by parts is not possible here.
I tried using a ...
1
vote
1answer
36 views
Further Stirling number series resummation
\begin{equation}
\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m
\end{equation}
Note: $S^{(3)}_m$ belongs to the Stirling number of the ...
0
votes
0answers
35 views
Stirling number series resummation
\begin{equation}\sum_{m=1}^{\infty}\frac{a_1^3 S_m^{(3)} (u-1)^{m-1}
\left(\frac{x}{x-1}\right)^m}{m!}\end{equation}
Does somebody know the result of this resummation?
Note:
$S_m^{(3)} $ belongs to ...
0
votes
0answers
76 views
General expression of a triangle sequence
\begin{gather*}
\frac{1}{4} \\
\frac{1}{4} \quad \frac{1}{4} \\
\frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\
\frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\
\frac{...
0
votes
1answer
48 views
General expression of a (maybe 3 or 2 dim) sequence [closed]
$\frac{1}{2}$
$\frac{1}{4}$ $\frac{1}{2}$
$\frac{1}{6}$ $\frac{1}{4}$ $\frac{11}{24}$
$\frac{1}{8}$ $\frac{1}{6}$ $\frac{11}{48}$ $\frac{5}{12}$
$\frac{1}{10}$ $\frac{1}{8}$ $\frac{11}{...
5
votes
1answer
110 views
How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$
I want to evaluate $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$
But I've not been successful in doing so, what I tried is
$$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\...
2
votes
1answer
92 views
Does the dilogarithm function (which is multi-valued) have a single-valued inverse?
The $p$-logarithm is defined for $|z|<1$ by
$$\text{Li}_p(z)=\sum_{n=1}^\infty\frac{z^n}{n^p}$$
and defined elsewhere in $\mathbb C$ by analytic continuation, though it may be multi-valued, ...
0
votes
0answers
34 views
Further question on Logarithm product integral
How to perform $\int_0^1 \frac{\left(a_0\log(u)+a_1\log(1-u)+a_{2}\log(1-xu)\right)^9}{u-1} du $?
Method tried:
Intgration-by-parts
Series expansion
change of variable $\log(u)=x$
But I still can't ...
1
vote
1answer
66 views
Powers of polylogarithms
I would like to take powers of arbitrary order to polylogarithm functions. For instance, given
$$
\text{Li}_\alpha(z) = \sum_{k=1}^\infty \frac{z^k}{k^\alpha}
$$
I am interested in
$$
[\text{Li}_\...
0
votes
0answers
9 views
Sub series as sub integral of lerch trascendent
If lerch trascendent is defined as follow
$\phi(z,s,a) = \sum_{k=0}^{\infty} \frac{z^{k}}{(a+k)^s} = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1} e^{-ax}}{(1-ze^{-x})}$ can we find the integral ...
11
votes
1answer
285 views
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ …
Question: How to show the relation
$$
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
$$
(using a "minimal industry" of relations, ...
1
vote
0answers
48 views
Upper bounds regarding polylogarithm $Li_p(e^w)$ when $|w| > 2\pi$ and p negative real
$\textit{The Computations of Polylogarithms, 1992,Technical report UKC, University of Kent, Canterbury, UK}\\ $ by $\textbf{David C Wood}$ says that
$$
Li_p(e^w) = \sum_{n\geq 0} \zeta(p-n)\frac{w^n}{...
5
votes
1answer
162 views
Integral from Mathematica's documentation: $\int_0^1 \frac{\log (\frac{1}{2}(1+\sqrt{4 x+1}))}{x} \, dx = \frac{\pi^2}{15} $
I like to peruse Mathematica's documentation and look at the 'Neat Examples': this is one I managed to figure out. Apparently it's due to Ramanujan:
$$
I=\int_0^1 \frac{\log \left(\frac{1}{2} \left(1+\...
21
votes
3answers
1k views
Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$
The following problem is proposed by a friend:
$$\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\{\...
9
votes
1answer
342 views
How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$.
How can i evaluate
$$\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$$
$$=\frac{\pi ^3}{6}-\frac{\pi ^3}{6\sqrt{2}}-4\pi +6\pi \ln \left(2\right)-\frac{\pi }...
2
votes
1answer
285 views
Evaluate $\Im(\operatorname{Li}_3(2i) + \operatorname{Li}_3(\frac i2))$
Applying the trilogarithm identity
$$ \operatorname{Li}_{3}\left(z\right) - \operatorname{Li}_{3}\left(1 \over z\right) =
-{1 \over 6}\ln^{3}\left(-z\right) -
{\pi^{2} \over 6}\ln\left(-z\right)\tag{1}...
2
votes
4answers
111 views
Errors are decreasing in series $\sum_{n=1}^\infty(-1)^n/n^4$?
Let $v=\sum_{n=1}^\infty(-1)^n/n^4$ ($v$ for "value"), let $S=(\sum_{n=1}^m(-1)^n/n^4)_{m\in\mathbb Z_{\ge1}}$ be the partial sums, and let $e=(|S_n-v|)_{n\in\mathbb Z_{\ge1}}$ be the errors....
2
votes
0answers
61 views
An integral involving a Gaussian and a power of a normal cumulative distribution function
Being inspired by How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$? we formulated the question below.
Let $c \in (0,1/\sqrt{2})$ and let $n \in \Bbb{N}$. Then let $\phi(x) : =\frac{\...
4
votes
0answers
204 views
Does there exist a closed form for $\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$?
I am not sure if there exists a closed form for
$$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$
which seems non-trivial.
I used the reflection and landen's identity, didn't help much.
...
4
votes
2answers
223 views
How to approach $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$ elegantly?
How to show that
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$
without using the generating function:
\...
5
votes
1answer
341 views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
14
votes
2answers
616 views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
10
votes
4answers
564 views
How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$
Before you think I haven't tried anything, please read.
I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
But I can't find a way to simplify it. ...
9
votes
3answers
454 views
How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way?
The following problem
$$\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)\tag1$$
was already ...
10
votes
1answer
269 views
Advanced Integral: $\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$
How to tackle
$$I=\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx\ ?$$
This integral came up while I was working on finding $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$.
First attempt: By ...
13
votes
3answers
508 views
How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$
Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$...
12
votes
1answer
661 views
How to evaluate $\int _0^1\frac{\ln \left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$ without the aid of trigonometric functions.
Earlier I made a post about the evaluation of a certain sum that had the following integral representation:
$$\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{\left(2k+1\right)^2}=\int _0^1\frac{\ln \...
5
votes
0answers
202 views
Is there a closed form without MZV for $ \sum _{k=1}^{\infty }\frac{H_k}{k^6\:2^k}$?
While evaluating the weight $7$ integral $\displaystyle \int_0^1\frac{\ln^3\left(1-x\right)\ln^3\left(1+x\right)}{1+x}\:dx$
I managed to prove that
$$\int_0^1\frac{\ln^3\left(1-x\right)\ln^3\left(1+x\...
9
votes
2answers
735 views
Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$
Here is a nice problem proposed by Cornel Valean
$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{...
4
votes
2answers
301 views
How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$
I want to evaluate
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$
Im not sure if this has a closed form, integration by parts is out of the question since there would be ...
4
votes
1answer
244 views
How can I evaluate $\int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
I am trying to evaluate $\displaystyle \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
I first tried using the series expansion for the dilogarithm like this
$$\sum _{n=1}^{\...
3
votes
1answer
75 views
On $\int_0^{2\pi }\frac{\prod_{k=1}^m \text{Li}_{a_k}(e^{-ix})-\prod_{k=1}^m \text{Li}_{a_k}(e^{ix})}{e^{-ix}-e^{ix}} \, dx$
OP of this post evaluated a lot of remarkable polylog integrals (without proof), from which I conjecture a generalized one ($a_k\in \mathbb N$):
$$\int_0^{2 \pi } \frac{\prod _{k=1}^m \text{Li}_{a_k}(...
33
votes
2answers
2k 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):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}...
1
vote
2answers
70 views
Closed-form expressions for the zeros of $\text{Li}_{-n}(x)$?
Consider the first few polylogarithm functions $\text{Li}_{-n}(x)$, where $-n$ is a negative integer and $x\in\mathbb R$ (plotted below). Observation suggests that $\text{Li}_{-1}(x)$ has one zero (at ...
1
vote
2answers
82 views
Evaluation of a log-trig integral in terms of the Clausen function (or other functions related to the dilogarithm)
Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(a,\theta\right)}:=\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2a\cos{\left(\...
0
votes
1answer
82 views
What's this partial sum $ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$ equal?
I want to get this partial sum of $$ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$$ which it is convergent and it is closed to one half , I have tried to use polylogarithm function which is defined as :...
5
votes
1answer
275 views
Evaluate $\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$
I encountered a hypergeometric integral while investigating harmonic sums
$$\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$$
Based on my experience I suspect a nice ...
2
votes
0answers
82 views
Evaluating $\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$ without using $\sum_{n=1}^\infty\frac{H_n}{n^3}x^n$
I am trying to evaluate
$$I=\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$$
Integration by parts yields
$$I=\frac58\zeta(4)-\frac12\int_0^1\frac{\ln(1-x)\text{Li}_2(-x^2)}{x}dx$$
Another related ...
3
votes
2answers
196 views
Is there a closed-form for $\sum_{n=0}^{\infty}\frac{n}{n^3+1}$?
I'm reading a book on complex variables (The Theory of Functions of a Complex Variable, Thorn 1953) and the following is shown:
Let $f(z)$ be holomorphic and single valued in $\mathbb{C}$ except at a ...
1
vote
1answer
82 views
How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?
where $H_n$ denotes the harmonic number.
I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even
$$-\sum_{n\geq 1}H_{2n}^2\...
2
votes
1answer
73 views
Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?
Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original ...
2
votes
1answer
92 views
Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.
I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of ...
