All Questions
43
questions
6
votes
1
answer
285
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
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
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
5
votes
0
answers
281
views
Is there a closed form without MZV for $ \sum _{k=1}^{\infty }\frac{H_k}{k^6\:2^k}$?
While evaluating the weight $7$ integral $\displaystyle \int_0^1\frac{\ln^3\left(1-x\right)\ln^3\left(1+x\right)}{1+x}\:dx$
I managed to prove that
$$\int_0^1\frac{\ln^3\left(1-x\right)\ln^3\left(1+x\...
- 2,646
4
votes
2
answers
411
views
How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$
I want to evaluate
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$
Im not sure if this has a closed form, integration by parts is out of the question since there would be ...
user805212
6
votes
1
answer
414
views
How can I evaluate $\int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
I am trying to evaluate $\displaystyle \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
I first tried using the series expansion for the dilogarithm like this
$$\sum _{n=1}^{\...
user806882
1
vote
1
answer
111
views
How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?
where $H_n$ denotes the harmonic number.
I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even
$$-\sum_{n\geq 1}H_{2n}^2\...
- 125
2
votes
1
answer
86
views
Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?
Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original ...
- 125
2
votes
1
answer
111
views
Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.
I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of ...
- 125
1
vote
2
answers
150
views
How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$
I am having a difficult time evaulating
$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$
I have tried the following relation:
$$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$
...
user753116
3
votes
2
answers
187
views
Closed form for $\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x$
I am looking for a closed form for:
$$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$
I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/...
user753116
1
vote
1
answer
256
views
Looking for closed form for $\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x$
I was evaluating and integral involving iterated logarithms when the following integral appeared:
$$\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \...
user753116
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
5
votes
3
answers
595
views
Closed form for the skew-harmonic sum $\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}$
In a post found here it is mentioned that a closed form for the so-called younger brother (younger in the sense the power in the denominator is only squared, rather than cubed as in the linked ...
- 11.8k
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