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Tagged with polylogarithm harmonic-numbers
99
questions
11
votes
0
answers
252
views
Solve the integral $\int_0^1 \frac{\ln^2(x+1)-\ln\left(\frac{2x}{x^2+1}\right)\ln x+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx$
I tried to solve this integral and got it, I showed firstly
$$\int_0^1 \frac{\ln^2(x+1)+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=2\Im\left[\text{Li}_3(1+i) \right] $$
and for other integral
$$\int_0^...
- 1,577
6
votes
1
answer
282
views
Calculate $\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$
this integral got posted on a mathematics group by a friend
$$I=\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$$
I tried seeing what I'd get from ...
- 231
1
vote
0
answers
94
views
Series with power of generalized harmonic number $\displaystyle\sum_{k=1}^{\infty}\left(H_k^{(s)}\right)^n x^k$
It's possible to generalize these series?
$$\sum_{k=1}^{\infty}H_k^{(s)}x^k=\frac{\operatorname{Li}_s(x)}{1-x}$$
$$\sum_{k=1}^{\infty}H_k^2 x^k=\frac{\ln(1-x)^2+\operatorname{Li}_2(x)}{1-x}$$
Where:
$$...
11
votes
0
answers
436
views
Is the closed form of $\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx$ known in the literature?
Using
$$\text{Li}_{2a+1}(x)-\text{Li}_{2a+1}(1/x)=\frac{i\,\pi\ln^{2a}(x)}{(2a)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\ln^{2k+1}(x)\tag{1}$$
and
$$\int_0^1x^{n-1}\operatorname{Li}_a(x)\mathrm{d}...
- 25.8k
35
votes
0
answers
2k
views
Are these generalizations known in the literature?
By using
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$
and
$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(...
- 25.8k
9
votes
1
answer
329
views
Different ways to evaluate $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$
The following question:
How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was ...
- 25.8k
8
votes
1
answer
342
views
Evaluating $\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}x\,dx$
Well, I've been trying to solve the following integral:
\begin{equation*}
\int_0^1\frac{\text{Li}_3(x)}{1+x}\mathrm dx,
\end{equation*}
where by integration by parts, making $u=\text{Li}_3(x)$ and $\...
- 481
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
7
votes
3
answers
280
views
Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$
I am trying to compute the following double sum
$$\boxed{\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)}$$
I proceeded as following
$$\sum_{n=1}^{\infty}\sum_{m=1}^...
- 2,811
3
votes
4
answers
475
views
How to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$
Edit
In this post I computed the following integral
$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)$$
Now I am trying to compute
$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}...
- 2,811
1
vote
1
answer
70
views
Evaluating $\sum_{k=1}^{a}\frac{-1-H_k}{k(1-e)^k}$
Question :
My attempt:
Let $a=17399172$
$$\begin{align}
&\sum_{k=1}^{a}\frac{-1-H_k}{\log_2\left(\sum_{j=0}^{k}\left(\ln\left(e^{C_j^k}\right)\right)\right)(1-e)^k} \\
&= \sum_{k=1}^{a}\frac{-...
- 2,200
6
votes
0
answers
305
views
Does there exist a closed form for $\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$?
I am not sure if there exists a closed form for
$$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$
which seems non-trivial.
I used the reflection and landen's identity, didn't help much.
...
- 25.8k
4
votes
2
answers
335
views
How to approach $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$ elegantly?
How to show that
$$\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)$$
without using the generating function:
\...
- 25.8k
6
votes
1
answer
765
views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
- 25.8k
17
votes
2
answers
1k
views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\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