All Questions
11
questions
6
votes
2
answers
326
views
How to show $\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G$
I am trying to prove that
$$\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G,$$
where $G$ is the Catalan constant and $\operatorname{Li}_2(x)$ is the dilogarithm ...
- 25.8k
3
votes
1
answer
314
views
Evaluating $\int_0^1\frac{\ln^2(1+x)+2\ln(x)\ln(1+x^2)}{1+x^2}dx$
How to show that
$$\int_0^1\frac{\ln^2(1+x)+2\ln(x)\ln(1+x^2)}{1+x^2}dx=\frac{5\pi^3}{64}+\frac{\pi}{16}\ln^2(2)-4\,\text{G}\ln(2)$$
without breaking up the integrand since we already know:
$$\int_0^1\...
- 25.8k
10
votes
1
answer
410
views
Proving $\int_0^{1/2}\frac{\text{Li}_2(-x)}{1-x}dx=-\text{Li}_3\left(-\frac12\right)-\frac{13}{24}\zeta(3)$
By comparing some results, I found that
$$\int_0^{\frac12}\frac{\text{Li}_2(-x)}{1-x}dx=-\text{Li}_3\left(-\frac12\right)-\frac{13}{24}\zeta(3).\tag{1}$$
I tried to prove it starting with applying IBP:...
- 25.8k
9
votes
1
answer
329
views
Different ways to evaluate $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$
The following question:
How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was ...
- 25.8k
4
votes
2
answers
180
views
Prove $\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$
I managed here to prove $$\int_0^1\frac{\text{Li}_2(-x^2)}{\sqrt{1-x^2}}\,dx=\pi\int_0^1\frac{\ln\left(\frac{2}{1+\sqrt{1+x}}\right)}{x}\,dx$$
but what I did was converting the LHS integral to a ...
- 25.8k
6
votes
1
answer
257
views
Integral from Mathematica's documentation: $\int_0^1 \frac{\log (\frac{1}{2}(1+\sqrt{4 x+1}))}{x} \, dx = \frac{\pi^2}{15} $
I like to peruse Mathematica's documentation and look at the 'Neat Examples': this is one I managed to figure out. Apparently it's due to Ramanujan:
$$
I=\int_0^1 \frac{\log \left(\frac{1}{2} \left(1+\...
- 8,369
4
votes
2
answers
335
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:
\...
- 25.8k
10
votes
4
answers
619
views
How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way?
The following problem
\begin{align}
&\int_{0}^{\pi/2}
x\ln^{2}\left(\sin\left(x\right)\right)\,{\rm d}x \\[5mm] = & \
\frac{1}{2}\ln^{2}\left(2\right)\zeta\left(2\right)
- \frac{19}{32}\,\zeta\...
- 25.8k
2
votes
0
answers
142
views
Evaluating $\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$ without using $\sum_{n=1}^\infty\frac{H_n}{n^3}x^n$
I am trying to evaluate
$$I=\int_0^1\frac{\ln(1+x^2)\text{Li}_2(x)}{x}dx$$
Integration by parts yields
$$I=\frac58\zeta(4)-\frac12\int_0^1\frac{\ln(1-x)\text{Li}_2(-x^2)}{x}dx$$
Another related ...
- 25.8k
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
$$I=-2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\...
- 25.8k
2
votes
2
answers
547
views
Compute $\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx$
How to prove
$$\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