All Questions
23
questions
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
8
votes
4
answers
688
views
How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$
Before you think I haven't tried anything, please read.
I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
But I can't find a way to simplify it. ...
user809806
9
votes
2
answers
941
views
Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$
Here is a nice problem proposed by Cornel Valean
$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{...
- 25.8k
2
votes
2
answers
241
views
Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$
How can we evaluate: $$\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\ ?$$
Where $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number.
I managed to find the closed form but had hard ...
- 25.8k
6
votes
5
answers
315
views
How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$
$$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$
This number looks like $\zeta(5)$ value.
We expand the terms
$$\int_0^1\frac{\frac{\pi^2}{...
- 1,343
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
10
votes
1
answer
991
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
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{2n + 1} = \frac{3 \pi}{...
- 11.8k
15
votes
3
answers
956
views
Compute $\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx$
How to evaluate $$\int_0^\infty \frac{\operatorname{Li}_3(x)}{1+x^2}\ dx\ ?$$
where $\displaystyle\operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}$ , $|x|\leq1$
I came across this integral ...
- 25.8k
6
votes
1
answer
308
views
On generalizing the harmonic sum $\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n = S_{k-1,2}(1)+\zeta(k+1)$ when $z=1$?
Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$. In this post it asks for the evaluation,
$$\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