All Questions
28
questions
4
votes
0
answers
220
views
How can we prove a closed form for $\frac{1}{8} \text{Li}_2\left(\frac{2+\sqrt{3}}{4} \right)+\text{Li}_2\left(2+\sqrt{3}\right)$?
I have been working on a problem in number theory that I have reduced to the problem of showing that the two-term linear combination
$$ \frac{1}{8} \text{Li}_2\left(\frac{2+\sqrt{3}}{4} \right) + \...
- 2,017
2
votes
1
answer
247
views
Finding a closed-form for the sum $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}$
Let $\mathcal{S}$ denote the sum of the following alternating series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}H_{2n}}{n^{4}}\approx-1.392562725547,$$
where $H_{n}$ denotes the $n$-...
- 30.7k
6
votes
0
answers
306
views
Does there exist a closed form for $\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$?
I am not sure if there exists a closed form for
$$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$
which seems non-trivial.
I used the reflection and landen's identity, didn't help much.
...
- 25.8k
44
votes
2
answers
3k
views
Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$
We have the following result ($\text{Li}_{n}$ being the polylogarithm):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}...
- 19.1k
5
votes
3
answers
320
views
Is there a closed-form for $\sum_{n=0}^{\infty}\frac{n}{n^3+1}$?
I'm reading a book on complex variables (The Theory of Functions of a Complex Variable, Thorn 1953) and the following is shown:
Let $f(z)$ be holomorphic and single valued in $\mathbb{C}$ except at a ...
- 8,369
2
votes
2
answers
241
views
Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$
How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$
Any help will be appreciated.
- 8,654
10
votes
3
answers
667
views
Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants
How can we evaluate
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$
Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by ...
- 8,654
7
votes
3
answers
282
views
The closed form for $\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n$
Is there a closed form for
$$\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n\ ?$$
Where $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number.
I managed to find the closed form but had hard ...
- 25.8k
16
votes
3
answers
918
views
How to compute $\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$?
Can we evaluate $\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$ ?
where $H_n=\sum_{k=1}^n\frac1n$ is the harmonic number.
A related integral is $\displaystyle\int_0^1\frac{\ln^2(1-x)\...
- 25.8k
14
votes
4
answers
2k
views
Compute $\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}dx$ or $\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$
Prove that
I encountered this integral while working on the sum $\displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$. Both of the integral and the sum were proposed by Cornel Valean:
The ...
- 25.8k
8
votes
1
answer
173
views
Expressing $\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 k\,2^k}$ as a finite sum involving $\zeta(\cdot)$, $Li_k(\cdot)$, $\pi$, and $\ln 2$
While working on the integral posted here, through a large amount of skulduggery, I managed to arrive at the following intriguing sum
$$\begin{align}\sum_{n = 1}^\infty \sum_{k = 1}^n \frac{1}{n^4 ...
- 11.8k
10
votes
1
answer
992
views
Computing $\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx$ or $\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}$
Challenging Integral:
\begin{align}
I=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx&=6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{16}\zeta(5)-\frac{21}{...
- 25.8k
2
votes
3
answers
187
views
Compute the integral in a closed form : $\int_0^{1}\operatorname{Li}_2(1-x)dx$
How I can find the closed form of the following integration :
$I=\int_0^{1}\operatorname{Li}_2(1-x)dx$
$J=\int_0^{1}\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-\frac{1}{x})dx$
$K=\int_0^{1}\ln(x)...
- 147
6
votes
1
answer
249
views
On the alternating Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1}$
In trying to evaluate the integral given here, in a rather circuitous way, I stumbled upon the following alternating Euler sum
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1} = \frac{3 \pi}{...
- 11.8k
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