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Tagged with polylogarithm harmonic-numbers
99
questions
1
vote
0
answers
144
views
Different approach to compute $\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{1+x}2\right)\ dx$
The following integral $$I=\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{1+x}2\right)\ dx$$
was already evaluated by @Knas here where he found
$$I=-2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\...
- 25.8k
3
votes
3
answers
204
views
Approaching $\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}$ elegantly
How to elegantly prove that $$\sum_{n=1}^\infty\frac{\overline{H}_n-H_{n/2}}{n^3}=2\text{Li}_4\left(\frac12\right)-\frac{49}{16}\zeta(4)+\frac72\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$
...
- 25.8k
7
votes
3
answers
282
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The closed form for $\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n$
Is there a closed form for
$$\sum_{n=1}^\infty \frac{H_{n/2}}{n^2}x^n\ ?$$
Where $H_{n/2}=\int_0^1\frac{1-x^{n/2}}{1-x}\ dx$ is the harmonic number.
I managed to find the closed form but had hard ...
- 25.8k
10
votes
2
answers
426
views
Is there a closed form function for the integral $\int_{0}^{z} \frac{1}{x}\log(1-x)\log(x)\log(1+x)\,dx$?
Recently, in connection with the problem of calculating generating functions of the antisymmetric harmonic number (https://math.stackexchange.com/a/3526006/198592, and What's the generating ...
- 12.7k
5
votes
3
answers
595
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Closed form for the skew-harmonic sum $\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}$
In a post found here it is mentioned that a closed form for the so-called younger brother (younger in the sense the power in the denominator is only squared, rather than cubed as in the linked ...
- 11.8k
10
votes
3
answers
478
views
Prove closed expression for $\int_0^1 \log(x) \log(1+x) \log(2+x)\,dx$
In Closed form of $\int_{0}^{1} \frac{\log(1+x)\log(2+x) \log(3+x)}{1+x}\,dx$ I have proposed an integral which I could not solve, and though there were some upvotes on the question no solution was ...
- 12.7k
6
votes
5
answers
315
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How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$
$$\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
5
votes
2
answers
452
views
Calculating $\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx$
How to prove in a simpe way that
$$\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx=\frac{13}{8}\ln2\zeta(2)-\frac{33}{32}\zeta(3)\ ?$$
where $\operatorname{Li}_2$ is the ...
- 25.8k
2
votes
2
answers
547
views
Compute $\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx$
How to prove
$$\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)...
- 25.8k
8
votes
2
answers
774
views
Challenging Sum: compute $\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)$
How to prove
$$\sum_{n=1}^\infty\frac{H_n}{2n+1}\left(\zeta(3)-H_n^{(3)}\right)=\frac74\zeta(2)\zeta(3)-\frac{279}{16}\zeta(5)+\frac43\ln^3(2)\zeta(2)-7\ln^2(2)\zeta(3)\\+\frac{53}4\ln(2)\zeta(4)-\...
- 25.8k
8
votes
2
answers
457
views
How to find $I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx$
$$I=\int_0^1\frac{\arctan^2x}{1+x}\left(\frac{\ln x}{1-x}+\ln(1+x)\right)dx=-\frac{\pi^4}{512}+\frac{3\pi^2}{128}\ln^22+\frac{\pi}{8}G\ln2-\frac{21}{64}\zeta(3)\ln2$$
This integral was proposed to me ...
- 5,507
11
votes
2
answers
605
views
Compute $\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx$
How to prove
$$I=\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac16\left(\frac{\pi^3}{12}-\pi\operatorname{Li}_2\left(\frac13\right)\right)$$
This problem is ...
- 25.8k
9
votes
1
answer
448
views
Compute $\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{\sqrt{1-x^2}}dx$
I need to prove
$$I=\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{\sqrt{1-x^2}}dx=\frac{5\pi}8\zeta(3)-\pi\ln2\zeta(2)+\pi\ln^32$$
to finish my solution for this probem.
What I did is the common ...
- 25.8k
2
votes
1
answer
403
views
Is there a closed form for $\int_0^1\frac{\ln^4(1-x)\operatorname{Li}_4(x)}{x}dx\ ?$
How to compute
$$I=\int_0^1\frac{\ln^4(1-x)\operatorname{Li}_4(x)}{x}dx\ ?$$
where $\operatorname{Li}_r$ is the polylogarithm function.
First I tried integration by parts but got complicated, so ...
- 25.8k
4
votes
1
answer
264
views
Compute $\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}$
How to prove, without calculating each sum separately that
$$\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=24\zeta(7)-4\zeta(2)\zeta(5)-15\zeta(3)\zeta(4)\ ?$$
...
- 25.8k