All Questions
Tagged with polylogarithm real-analysis
29
questions
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:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{...
- 11.6k
20
votes
4
answers
1k
views
Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
- 5,626
17
votes
2
answers
894
views
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...
Question: How to show the relation
$$
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
$$
(using a "minimal industry" of relations, ...
- 34.2k
6
votes
1
answer
765
views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
- 25.8k
6
votes
3
answers
691
views
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?
I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
- 25.8k
16
votes
5
answers
1k
views
Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $
I proved the following result
$$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$
After ...
- 14.4k
4
votes
3
answers
686
views
Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$
This problem was already solved here (in different closed form).
But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $
...
- 25.8k
23
votes
3
answers
2k
views
Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$
The following problem is proposed by a friend:
$$\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\{\...
- 25.8k
17
votes
2
answers
1k
views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
- 25.8k
11
votes
2
answers
728
views
How to compute $\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$ or $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$
How to tackle
$$I=\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx\ ?$$
This integral came up while I was working on finding $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$.
First attempt: By ...
- 25.8k
11
votes
2
answers
927
views
Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$
where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$
these two sums are already solved by Cornel using ...
- 25.8k
10
votes
1
answer
790
views
A generalized "Rare" integral involving $\operatorname{Li}_3$
In my previous post, it can be shown that
$$\int_{0}^{1}
\frac{\operatorname{Li}_2(-x)-
\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x)
-\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} }
=\...
- 5,370
6
votes
1
answer
494
views
Is the closed form of $\int_0^1 \frac{x\ln^a(1+x)}{1+x^2}dx$ known in the literature?
We know how hard these integrals
$$\int_0^1 \frac{x\ln(1+x)}{1+x^2}dx;
\int_0^1 \frac{x\ln^2(1+x)}{1+x^2}dx;
\int_0^1 \frac{x\ln^3(1+x)}{1+x^2}dx;
...$$
can be. So I decided to come up with a ...
- 25.8k
1
vote
1
answer
164
views
Integral involving product of dilogarithm and an exponential
I am interested in the integral
\begin{equation}
\int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast)
\end{equation}
where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to ...
- 753
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(...
- 19.4k