All Questions
Tagged with polylogarithm logarithms
70
questions
12
votes
3
answers
460
views
How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$
I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ ...
- 5,370
4
votes
3
answers
211
views
Does $\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)$ have a closed form?
I am wondering if anyone has a nice way of approaching the following definite integral $\newcommand{\dilog}{\operatorname{Li}_2}$
$$\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\...
- 179
5
votes
2
answers
368
views
Closed form of the sum $s_4 = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^4}$
I am interested to know if the following sum has a closed form
$$s_4 = \sum_{n=1}^{\infty}(-1)^n \frac{H_{n}}{(2n+1)^4}\tag{1}$$
I stumbled on this question while studying a very useful book about ...
- 12.7k
4
votes
1
answer
82
views
Evaluate the following series sum.
Problem
I’m trying to evaluate the following series sum
\begin{equation}
S_{j}(z) = \sum_{k=1}^{\infty} \frac{2 H_{k} z^{k+2}}{(k+1)(k+2)^{j}}
\end{equation}
Where
\begin{equation}
H_{k} = \sum_{n=1}^{...
- 87
1
vote
1
answer
83
views
Integral of a modified softplus function
In a manuscript I am currently reading, the authors propose a modified softplus function
$$g(a)=\frac{\log\left(2^a +1 \right)}{\log(2)}$$
for some $a \in \mathbb{R}$. The authors then claim that if $...
- 961
4
votes
0
answers
341
views
How to evaluate $\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$
I am trying to evaluate
$$\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$$
But I am not sure what to do since integration by parts is not possible here.
I tried using a ...
user809806
1
vote
1
answer
59
views
Further Stirling number series resummation
\begin{equation}
\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m
\end{equation}
Note: $S^{(3)}_m$ belongs to the Stirling number of the ...
- 99
0
votes
0
answers
59
views
Stirling number series resummation
\begin{equation}\sum_{m=1}^{\infty}\frac{a_1^3 S_m^{(3)} (u-1)^{m-1}
\left(\frac{x}{x-1}\right)^m}{m!}\end{equation}
Does somebody know the result of this resummation?
Note:
$S_m^{(3)} $ belongs to ...
- 99
0
votes
0
answers
82
views
General expression of a triangle sequence
\begin{gather*}
\frac{1}{4} \\
\frac{1}{4} \quad \frac{1}{4} \\
\frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\
\frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\
\frac{...
- 99
0
votes
1
answer
53
views
General expression of a (maybe 3 or 2 dim) sequence [closed]
$\frac{1}{2}$
$\frac{1}{4}$ $\frac{1}{2}$
$\frac{1}{6}$ $\frac{1}{4}$ $\frac{11}{24}$
$\frac{1}{8}$ $\frac{1}{6}$ $\frac{11}{48}$ $\frac{5}{12}$
$\frac{1}{10}$ $\frac{1}{8}$ $\frac{11}{...
- 99
5
votes
1
answer
223
views
How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$
I want to evaluate $$\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)\...
user809806
0
votes
0
answers
50
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 ...
- 99
0
votes
1
answer
97
views
dilogarithm property.
Prove that
$\mathrm{Li}_{2}(-z)+\mathrm{Li}_{2}\left(\frac{z}{1+z}\right)=-\frac{1}{2} \ln ^{2}(1+z)$
I tried to paint in the rows, but I did not succeed. I don't have any more ideas.
- 222
1
vote
1
answer
101
views
Integral $\int_{0}^{e} \frac{\operatorname{W(x)} - x}{\operatorname{W(x)} + x} dx$
$$\int_{0}^{e} \frac{\operatorname{W(x)} - x}{\operatorname{W(x)} + x} dx = 2 \operatorname{Li_2(-e)} - e + \frac{\pi^2}{6} - \log(4) + 4 \log(1 + e)≈-0.819168$$
As usual I prefer to know if there is ...
0
votes
0
answers
61
views
How can we show that the multi-polylogarithmic function $L_{\underbrace{1,\ldots,1}_n}(z)=\frac{1}{n!}(L_1(z))^n$
How can we show that the multi-polylogarithmic function
$$L_{\underbrace{1,\ldots,1}_n}(z)=\frac{1}{n!}(L_1(z))^n.$$
Here $L_1(z)=-log(1-z)$.
I know that $\frac{d}{dz}L_{k_1,\ldots,k_r}(z)=\frac{1}{1-...
- 517