All Questions
22 questions
17
votes
3
answers
671
views
Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$
I was able to find
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x=-\frac14\sum_{n=1}^\infty\frac{4^n}{{2n\choose n}}\frac{H_{2n}}{n^3}$$
$$=5\operatorname{Li}_4\left(\frac12\right)-\frac{...
- 26.6k
7
votes
0
answers
114
views
Convergence of double integral involving logarithms and binomials? Can anyone present a simpler proof?
I am working through a paper I posted to arXiv and am looking to condense my results as well as seek alternative proofs for some of my intermediate results. In particular, I am looking for a shorter ...
- 6,279
9
votes
1
answer
334
views
Different ways to evaluate $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$
The following question:
How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was ...
- 26.6k
9
votes
4
answers
348
views
Alternative ways to evaluate $\int_0^1 x^{2n-1}\ln(1+x)dx =\frac{H_{2n}-H_n}{2n}$
For all, $n\geq 1$, prove that
$$\int_0^1 x^{2n-1}\ln(1+x)\,dx =\frac{H_{2n}-H_n}{2n}\tag1$$
This identity I came across to know from here, YouTube which is proved in elementary way.
Trying to make ...
- 3,482
4
votes
2
answers
349
views
How to approach $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$ elegantly?
How to show that
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$
without using the generating function:
\...
- 26.6k
6
votes
2
answers
386
views
Computing $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{2n+1}$ in an alternative way
The following equality
$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{2n+1}=\frac{3}{16}\pi^3-\frac34\ln^2(2)\pi-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
can be proved if we are allowed ...
- 26.6k
3
votes
3
answers
216
views
Alternative proof of computing $\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}$
In this solution we showed that
$$\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag1$$
using the identity
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\...
- 26.6k
1
vote
0
answers
147
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
$$I=-2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\...
- 26.6k
2
votes
4
answers
278
views
Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$
How to prove that
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$$
$$=2\text{Li}_4\left(\frac12\right)-2\zeta(4)+\frac{15}8\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)$$
where $\...
- 26.6k
3
votes
0
answers
201
views
Evaluating $\int_0^1\frac{\ln x\operatorname{Li}_2(x)\ln(1-x/2)}{x}\ dx$
How to evaluate
$$\int_0^1\frac{\ln x\operatorname{Li}_2(x)\ln(1-\frac{x}{2})}{x}\ dx\ ?$$
I came across this integral while I was trying to calculate $\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{...
- 26.6k
5
votes
0
answers
151
views
Computing $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^4}$
I managed to find
$$S=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^4}=20\beta(6)+\frac12\zeta(2)\beta(4)-\frac{7\pi^3}{32}\zeta(3)-\frac{31\pi}{8}\zeta(5)$$
where $\beta(a)$ is the ...
- 26.6k
2
votes
2
answers
182
views
Computing $\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}$
How to elegantly prove that
$$S=\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}=3\operatorname{Li}_4\left(\frac12\right)-\frac{29}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac18\ln^42$$
where $\...
- 26.6k
10
votes
2
answers
453
views
Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$
How to prove that
$$S=\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
I came up ...
- 26.6k
4
votes
2
answers
358
views
Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}$
how to prove that
$$\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
This problem ...
- 26.6k
2
votes
2
answers
566
views
Compute $\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx$
How to prove
$$\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)...
- 26.6k
14
votes
2
answers
511
views
An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$
The following classical generalization
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$
where $\eta(a)=\...
- 26.6k
6
votes
2
answers
483
views
How to compute $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ by real integration only?
How to prove, by real methods that
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$
where $H_n$ is the harmonic number and $\zeta$ is the Riemann zeta ...
- 26.6k
4
votes
0
answers
215
views
Compute $\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx$ using harmonic series
How to prove
$$\mathcal{I}=\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2(2)$$
This problem is proposed by a friend and I managed to compute it using only ...
- 26.6k
1
vote
2
answers
219
views
How to compute $\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}$?
How to prove
$$\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=\frac18\zeta(2)\zeta(3)-\frac{25}{32}\zeta(5)?$$
I came across this series while I was working on a nice integral $\int_0^1\frac{\ln(1+x)\...
- 26.6k
12
votes
5
answers
720
views
Prove $\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$
How to prove
$$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
I used this identity to solve some advanced harmonic series but I didn't provide a proof so I see that it's worth a post so that we ...
- 26.6k
3
votes
1
answer
378
views
Advanced : Compute $\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}$
How to prove the following equality
$$\mathcal S=\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}\\=672\zeta(9)-240\zeta(2)\zeta(7)-105\zeta(3)\...
- 26.6k
8
votes
1
answer
697
views
Resistant integral $\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$
Prove, without using harmonic series, that
$$I=\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$$
$$=\frac18\zeta(5)-\frac12\ln2\zeta(4)+2\ln^22\zeta(3)-\...
- 26.6k