All Questions
7 questions
8
votes
2
answers
476
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 ...
- 5,617
10
votes
2
answers
849
views
Prove $\zeta(3)=2\sum_{n=1}^\infty\frac{H_n}{n}\left[\frac1{4^n}{2n\choose n}\left(H_{2n}-H_n-\frac1{2n}-\ln2\right)+\frac1{2n}\right]$
How to prove
$$\zeta(3)=2\sum_{n=1}^\infty\frac{H_n}{n}\left[\frac1{4^n}{2n\choose n}\left(H_{2n}-H_n-\frac1{2n}-\ln2\right)+\frac1{2n}\right]$$
where $H_n$ is the harmonic number and $\zeta$ is ...
- 26.6k
1
vote
1
answer
146
views
Calulating $\int_0^1\frac{5\ln^4x\ln^4(1-x)-8\ln^5x\ln^3(1-x)+4\ln^2(1-x)\ln^6x}{x(1-x)}\ dx$
How to prove without using the derivative of Beta function that
$$\int_0^1\frac{5\ln^4(1-x)\ln^4x-8\ln^3(1-x)\ln^5x+4\ln^2(1-x)\ln^6x}{x(1-x)}\ dx\\=5760\left\{4\zeta(9)-\zeta(2)\zeta(7)-\zeta(4)\...
- 26.6k
5
votes
2
answers
363
views
Compute $\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^3}$
How to prove
$$\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^3}=\frac{31}{8}\zeta(5)-\frac{45}{8}\ln2\zeta(4)+\frac72\ln^22\zeta(3)-\frac78\zeta(2)\zeta(3)$$
where $H_n$ is the harmonic number and $\...
- 26.6k
7
votes
1
answer
472
views
Compute $\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}$
How to prove that
$$\small{\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n+1)^3}=\frac1{12}\ln^52+\frac{13}{128}\zeta(5)-\frac12\ln^32\zeta(2)+\frac74\ln^22\zeta(3)-\frac{17}{8}\ln2\zeta(4)+2\ln2\...
- 26.6k
20
votes
3
answers
1k
views
Compute $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^3}$
How to prove
$$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^3}=\frac{19}{2}\zeta(3)\zeta(4)-2\zeta(2)\zeta(5)-7\zeta(7)\ ?$$
where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th ...
- 26.6k
4
votes
2
answers
528
views
Compute $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}$ and $\sum_{n=1}^\infty\frac{H_n^2}{n^7}$
How to prove that
$$S_1=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=7\zeta(2)\zeta(7)+2\zeta(3)\zeta(6)+4\zeta(4)\zeta(5)-\frac{35}{2}\zeta(9)\ ?$$
$$S_2=\sum_{n=1}^\infty\frac{H_n^2}{n^7}=-\zeta(2)\...
- 26.6k