All Questions
Tagged with polylogarithm real-analysis
60
questions
30
votes
3
answers
921
views
A closed form for a lot of integrals on the logarithm
One problem that has been bugging me all this summer is as follows:
a) Calculate
$$I_3=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \ln{(1-x)} \ln{(1-xy)} \ln{(1-xyz)} \,\mathrm{d}x\, \mathrm{d}y\, \mathrm{...
- 4,044
29
votes
7
answers
1k
views
Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$
I have spent my holiday on Sunday to crack several integral & series problems and I am having trouble to prove the following integral
\begin{equation}
\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(...
- 19.4k
23
votes
3
answers
2k
views
Challenging problem: Calculate $\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$
The following problem is proposed by a friend:
$$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$
$$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\...
- 25.8k
22
votes
4
answers
1k
views
Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$?
Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\frac{1}{9}\right) &=\frac{{...
- 11.6k
20
votes
4
answers
1k
views
Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$
How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
- 5,626
19
votes
3
answers
709
views
Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$
It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about
the cubic version, that is
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$
What are ...
- 44.4k
17
votes
1
answer
1k
views
A rare integral involving $\operatorname{Li}_2$
A rare but interesting integral problem:
$$\int_{0}^{1}
\frac{\operatorname{Li}_2(-x)-
\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x)
-\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} }
=\...
- 5,370
17
votes
2
answers
1k
views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
- 25.8k
17
votes
2
answers
894
views
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...
Question: How to show the relation
$$
J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt
=-\frac 1{64}\pi^4
$$
(using a "minimal industry" of relations, ...
- 34.2k
16
votes
5
answers
1k
views
Double Euler sum $ \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} $
I proved the following result
$$\displaystyle \sum_{k\geq 1} \frac{H_k^{(2)} H_k}{k^3} =- \frac{97}{12} \zeta(6)+\frac{7}{4}\zeta(4)\zeta(2) + \frac{5}{2}\zeta(3)^2+\frac{2}{3}\zeta(2)^3$$
After ...
- 14.4k
16
votes
2
answers
889
views
Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $
Evaluate :
$$ \int_{0}^{1} \log\left(\dfrac{x^2-2x-4}{x^2+2x-4}\right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} $$
Introduction : I have a friend on another math platform who proposed a summation ...
- 5,558
13
votes
3
answers
689
views
How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$
Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$...
- 25.8k
12
votes
2
answers
718
views
General expressions for $\mathcal{L}(n)=\int_{0}^{\infty}\operatorname{Ci}(x)^n\text{d}x$
Define $$\operatorname{Ci}(x)=-\int_{x}^{
\infty} \frac{\cos(y)}{y}\text{d}y.$$
It is easy to show
$$
\mathcal{L}(1)=\int_{0}^{\infty}\operatorname{Ci}(x)\text{d}x=0
$$
and
$$\mathcal{L}(2)=\int_{0}^{\...
- 5,370
11
votes
2
answers
927
views
Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$
where $H_n$ is the harmonic number and can be defined as:
$H_n=1+\frac12+\frac13+...+\frac1n$
$H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$
these two sums are already solved by Cornel using ...
- 25.8k
11
votes
3
answers
561
views
Is the following Harmonic Number Identity true?
Is the following identity true?
$$ \sum_{n=1}^\infty \frac{H_nx^n}{n^3} = \frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(...
- 5,558