All Questions
24
questions
1
vote
1
answer
152
views
How to evaluate $\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{4}}$
It may be rather tedious and I will have to delve into deeper, but I have a little something. We probably already know this one. The thing is, the first one results in yet another Euler sum. But, I ...
10
votes
2
answers
635
views
Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$
While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried ...
- 25.8k
3
votes
1
answer
243
views
Is there closed form for $\sum_{k=1}^\infty (-1)^k\frac{H_k^{(4)}}{k^2}$?
Do we have closed form for
$$\sum_{k=1}^\infty (-1)^k\frac{H_k^{(4)}}{k^2}\ ?$$
where $H_k$ is the harmonic number.
I encountered this sum while I was working on calculating $\displaystyle \sum_{...
- 25.8k
5
votes
1
answer
198
views
Is there closed form for $\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^3}\ ?$
Is it possible to compute
$$\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^3}\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number and $H_n=\int_0^1\...
- 25.8k
7
votes
2
answers
339
views
Evaluating $\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$
I ma trying to prove
$$S=\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\...
- 25.8k
12
votes
3
answers
455
views
Computing $\int_0^1\frac{\ln(1-x^2)}{x}\operatorname{Li}_2\left(\frac{1-x}{2}\right)\ dx$
How to evaluate
$$I=\int_0^1\frac{\ln(1-x^2)}{x}\operatorname{Li}_2\left(\frac{1-x}{2}\right)\ dx\ ?$$
This integral was mentioned by @nospoon in the comments of this problem.
What I tried is ...
- 25.8k
5
votes
1
answer
328
views
Integral $\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\ln2\right)\ dx$
I am trying to evaluate
$$I=\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\ln2\right)\ dx$$
I encountered this integral while I was trying to calculate the integral
$$\...
- 25.8k
4
votes
1
answer
202
views
Seeking to evaluate: $\int_{0}^{1}\arctan(x)\cdot\ln\frac{x+x^3}{(1-x)^2}\cdot\frac{\mathrm dx}{x}$
We wish to evaluate this integral,
$$\int_{0}^{1}\arctan(x)\cdot\ln\frac{x+x^3}{(1-x)^2}\cdot\frac{\mathrm dx}{x}$$
I have tried using substitution, $u=\frac{x+x^3}{(1-x)^2}$ and integration by parts ...
- 1,343
9
votes
1
answer
331
views
Evaluating $\int_0^{\pi/2} \frac{t \ln (1-\sin{t})}{\sin t} dt$
In a problem in scattering theory, this integral arises:
$$\displaystyle{\int\limits_0^{\pi/2} \frac{t \ln (1-\sin{t})}{\sin t} dt}$$ I have tried a number of approaches to evaluating the integral, ...
- 176
8
votes
5
answers
357
views
Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$
How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \...
- 25.8k
3
votes
1
answer
356
views
Calculate $\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx$
Prove that
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{\pi^3}{32}$$
I managed to prove the above equality using integral ...
- 25.8k
6
votes
3
answers
745
views
Compute $ \int_0^1\frac{\ln^2(1+x)}{1+x^2}\, dx$
How to prove
$$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$
Where $ \operatorname{Li}_3(x)$ is the the trilogarithm ...
- 25.8k
5
votes
3
answers
526
views
Evaluate $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}dx$
How to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x^2}\ dx=\text{Im}\left(\operatorname{Li}_3(1+i)\right)-\frac{\pi^3}{32}-G\ln2 \ ?$$
where $\operatorname{Li}_3(x)=\sum\limits_{n=1}^\infty\frac{x^n}{...
- 25.8k
11
votes
3
answers
786
views
How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$
How to evaluate $$\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$$
I tried to integrate by parts, but no way so far, help me, thanks.
- 731
8
votes
3
answers
711
views
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$
I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches ...
- 25.8k