All Questions
Tagged with polylogarithm harmonic-numbers
99
questions
2
votes
3
answers
187
views
Compute the integral in a closed form : $\int_0^{1}\operatorname{Li}_2(1-x)dx$
How I can find the closed form of the following integration :
$I=\int_0^{1}\operatorname{Li}_2(1-x)dx$
$J=\int_0^{1}\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-\frac{1}{x})dx$
$K=\int_0^{1}\ln(x)...
- 147
1
vote
0
answers
119
views
Generalized form of this Harmonic Number series $\sum_{n=1}^{\infty} \frac{{H_n}x^{n+1}}{(n+1)^3}$
i've tried to Evaluate $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx$$ without using Contour integral
first i changed $2\sin(x)$ into polar form ,and i got $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx ...
- 347
6
votes
1
answer
249
views
On the alternating Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1}$
In trying to evaluate the integral given here, in a rather circuitous way, I stumbled upon the following alternating Euler sum
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1} = \frac{3 \pi}{...
- 11.8k
8
votes
2
answers
1k
views
Evaluate $\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx$
How to prove $\ \displaystyle\int_0^1\frac{\ln x\ln(1-x)}{1+x^2}\ dx=\frac{3\pi^3}{64}+\frac{\pi}{16}\ln^22-\Im\operatorname{Li}_3(1+i)$
Where $\ \displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\...
- 25.8k
5
votes
2
answers
471
views
Integral $\ 4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx$
How to prove that
$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$
Where $\chi_2(x)=\sum_{...
- 25.8k
15
votes
3
answers
956
views
Compute $\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$
How to evaluate $$\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ ?$$
where $\displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}$ , $|x|\leq1$
I came across this integral ...
- 25.8k
4
votes
2
answers
624
views
challenging sum $\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}$
How to prove that
\begin{align}
\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)
\end{align}
...
- 25.8k
4
votes
2
answers
503
views
Sum $\sum\limits_{n=1}^\infty\frac{H_n^2}{n^22^n}$
Where $ H_n$ is the harmonic number, $\ \displaystyle H_n=1+\frac12+\frac13+...+\frac1n$.
I am going to present my solution as I need it as a reference.
Other approaches are appreciated.
here is ...
- 25.8k
6
votes
2
answers
529
views
$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$
This problem was proposed by Cornel and he showed that
$$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+...
- 25.8k
1
vote
1
answer
549
views
Find the closed form of $\quad\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}$
where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(3)}=1+\frac1{2^3}+\frac1{3^3}+...+\frac1{n^3}$
I managed to prove
$\quad\displaystyle\sum_{n=1}^...
- 25.8k
11
votes
2
answers
927
views
Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$
where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$
these two sums are already solved by Cornel using ...
- 25.8k
1
vote
1
answer
159
views
Closed form for $f(z,a)=\sum_{n=1}^\infty \frac{H_n}{n^a}z^n$?
So I have been working with some power series involving the Harmonic numbers. I have been able to evaluate the first couple of sums as
$$f(z,0)=\sum_{n=1}^\infty H_n z^n=\sum_{n=1}^\infty z^n\int_0^1 \...
- 4,027
6
votes
1
answer
308
views
On generalizing the harmonic sum $\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n = S_{k-1,2}(1)+\zeta(k+1)$ when $z=1$?
Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$. In this post it asks for the evaluation,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\tfrac{5}{4}\zeta(4)$$
while this post and this ...
- 54.9k
3
votes
5
answers
290
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:
$$\...
user668815
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