All Questions
Tagged with polylogarithm summation
23
questions
2
votes
2
answers
222
views
+100
$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$ as a limit of a sum
Working on the same lines as
This/This and
This
I got the following expression for the Dilogarithm $\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$:
$$\operatorname{Li}_{2} \left(\frac{1}{e^{\...
- 814
0
votes
0
answers
88
views
$\operatorname{Li}_{2} \left(\frac12 \right)$ vs $\operatorname{Li}_{2} \left(-\frac12 \right)$ : some long summation expressions
Throughout this post, $\operatorname{Li}_{2}(x)$ refers to Dilogarithm.
While playing with some Fourier Transforms, I came up with the following expressions:
$$2 \operatorname{Li}_{2}\left(\frac12 \...
- 814
4
votes
1
answer
224
views
Closed form for $\sum\limits_{M=2}^{\infty}\sum\limits_{n=M}^{\infty} \frac{2^{-M}3^{M-n-1}\pi \csc(n \pi) \Gamma(M)}{(n+1)^2 \Gamma(M-n)\Gamma(n+1)}$
I encountered this expression generated by mathematica as a sub-step in a problem I am solving.
$$\sum\limits_{M=2}^{\infty}\sum\limits_{n=M}^{\infty} \frac{2^{-M}3^{M-n-1}\pi \csc(n \pi) \Gamma(M)}{(...
- 814
3
votes
1
answer
116
views
Is there a closed form solution for the sum $\sum\limits_{M=2}^{\infty} \sum\limits_{n=M}^{\infty} \frac{2^{-M}3^{M-n-1}}{(M-n-1)(n+1)^{2}}$?
While working on another problem, this came up as a sub-step:
$$
\sum_{M\ =\ 2}^{\infty}\,\,\,\sum_{n\ =\ M}^{\infty}\
{2^{-M}\ 3^{M - n - 1} \over \left(M - n - 1\right)\left(n + 1\right)^{\,2}}
$$
...
- 814
2
votes
0
answers
140
views
The ultimate polylogarithm ladder
As you can see, here I performed a derivation of a quite simple formula, not much differing from the standard integral representation of the Polylogarithm. Seeking to make it fancier, I arrived at ...
- 1,039
1
vote
0
answers
68
views
Polylogarithm further generalized
Here I proposed a generalized formula for the polylogarithm. However, because of a slight mistake towards the end, visible prior to the edit, I was unaware that it yields just a result of an integral ...
- 1,039
2
votes
1
answer
255
views
Generalized formula for the polylogarithm
Some time ago, I discovered the formula for repeated application of $z\frac{d}{dz}$ here. Recently, I thought about taking the function to which this would be applied to be the integral representation ...
- 1,039
0
votes
1
answer
98
views
Converting a 2D lattice sum into a sum over 1D lattice sums in a circle
I'm working on a physics problem. I have a lattice sum, which in 1D is a sum over a linear chain. It reads
$$
f_k = \sum_{n=1}^\infty \frac{2\cos(kn)}{n^3}.
$$
This can be written in terms of ...
- 510
2
votes
0
answers
75
views
The summation $\sum_{k=1}^{\infty} \frac{\phi^{-2k}}{k^2}=\text{Li}_2(1/\phi^2)=\frac{\pi^2}{15}-\ln^2 \phi$ [duplicate]
In a recent post
Computing $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$, $\zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2}$
I evaluated the required ...
- 43.5k
3
votes
3
answers
392
views
Logarithmic integral $ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\ \mathrm{d}x $
I found this integral weeks ago.
$$ \int_0^1 \dfrac{x\ln(x)\ln(1+x)}{1+x^2}\ \mathrm{d}x $$
I tried to solve this integral using various series representation and ended up with a complicated double ...
- 2,139
1
vote
1
answer
59
views
Further Stirling number series resummation
\begin{equation}
\sum_{m=1}^\infty\sum_{n=1}^\infty (-1)^{n } \frac{S_m^{(3)}}{m! n}(-1 + u)^{(m + n - 1)} (\frac{x}{-1 + x})^m
\end{equation}
Note: $S^{(3)}_m$ belongs to the Stirling number of the ...
- 99
0
votes
0
answers
59
views
Stirling number series resummation
\begin{equation}\sum_{m=1}^{\infty}\frac{a_1^3 S_m^{(3)} (u-1)^{m-1}
\left(\frac{x}{x-1}\right)^m}{m!}\end{equation}
Does somebody know the result of this resummation?
Note:
$S_m^{(3)} $ belongs to ...
- 99
0
votes
0
answers
82
views
General expression of a triangle sequence
\begin{gather*}
\frac{1}{4} \\
\frac{1}{4} \quad \frac{1}{4} \\
\frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\
\frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\
\frac{...
- 99
0
votes
1
answer
101
views
What's this partial sum $ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$ equal?
I want to get this partial sum of $$ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$$ which it is convergent and it is closed to one half , I have tried to use polylogarithm function which is defined as :...
- 7,185
32
votes
1
answer
817
views
On the relationship between $\Re\operatorname{Li}_n(1+i)$ and $\operatorname{Li}_n(1/2)$ when $n\ge5$
Motivation
$\newcommand{Li}{\operatorname{Li}}$
It is already known that:
$$\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, ...
- 8,669