All Questions
Tagged with polylogarithm real-analysis
60
questions
4
votes
1
answer
439
views
Advanced Sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}$
How to prove
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}=
\\ \small{\frac43\ln^32\zeta(2)-\frac72\ln^22\zeta(3)-\frac{21}{16}\zeta(2)\zeta(3)+\frac{713}{64}\zeta(5)-\frac4{15}\ln^52-8\ln2\...
- 25.8k
1
vote
0
answers
119
views
Generalized form of this Harmonic Number series $\sum_{n=1}^{\infty} \frac{{H_n}x^{n+1}}{(n+1)^3}$
i've tried to Evaluate $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx$$ without using Contour integral
first i changed $2\sin(x)$ into polar form ,and i got $$\int_{0}^{\frac{\pi}{6}}x\ln^2(2\sin(x))dx ...
- 347
8
votes
0
answers
404
views
Powerful Integral $\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt$
This integral can be found in Cornel's book, (Almost) Impossible Integral, Sums and Series page $97$ where he showed that
$$\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt=\frac14\left(\frac12\ln^2(1+x^2)-2\...
- 25.8k
4
votes
3
answers
686
views
Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$
This problem was already solved here (in different closed form).
But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $
...
- 25.8k
6
votes
2
answers
529
views
$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$
This problem was proposed by Cornel and he showed that
$$S=\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac13\ln^42-\frac12\ln^22\zeta(2)+\frac72\ln2\zeta(3)-\frac{21}4\zeta(4)+...
- 25.8k
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
6
votes
3
answers
691
views
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?
I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
- 25.8k
6
votes
1
answer
181
views
Series power function over exponential function
A typical exercise from calculus is to show that any exponential function eventually grows faster than any power function, i.e.
$$ \lim_{k \to \infty} \frac{k^a}{b^k} = 0 \qquad \text{ for } a,b>1.$...
- 161
2
votes
0
answers
368
views
Upper bound the Polylogarithm $\sum_{n=1}^\infty \frac{x^n}{n^2}$
Let $x \in (0,1)$ be some real number, we can then consider the Polylogarithm:
$$\operatorname{L}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$$
It is not hard to see that the following upper bound holds:
$$...
- 1,135
0
votes
0
answers
42
views
Upper-bounding $\exp \log^{d} \frac{\log n}{n}$
How would you upper-bound this expression?
$$f(n, d) = \exp \log^{d} \frac{\log n}{n}$$
If $d = 1 $ this woulld simplify to $\frac{\log n}{n}$.
Any suggestions on how to upperbound it?
Notation ...
- 2,670
5
votes
1
answer
583
views
Polylogarithms: How to prove the asympotic expression $ z \le \mathrm{Li}_{s}(z) \le z(1+2z 2^{-s}), \;z<-1, \;s \gg \log_2|z|$
For $|z| < 1, s > 0$ the polylogarithm has the power series
$$\mathrm{Li}_{s}(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots = z\left(1+ {z \over 2^s} + {...
- 18.9k
9
votes
1
answer
218
views
On the monotony of $-\int_0^1\frac{e^y}{y}(\operatorname{Li}_x(1-y)-\zeta(x)) \,\mathrm dy$
$\def\d{\mathrm{d}}$After I was studying variations of the integral representation for $\zeta(3)$ due to Beukers, see the section More complicated formulas from this Wikipedia, I've thought an ...
user243301
3
votes
0
answers
106
views
Proving that swapping the order of this summation is justified
I'm unsure if this has been discovered already, but it's heavily related to my current research, particularly to this question of mine (this conjecture was also originally posted at the beginning of ...
- 856
2
votes
0
answers
85
views
Natural proof of identity for $\text{Li}_2(x)+\text{Li}_2(1-x)$ [duplicate]
What's a natural way to compute $\text{Li}_2(x)+\text{Li}_2(1-x)$ in closed form ?
Once you know the answer $\text{Li}_2(x)+\text{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x)$ , computing the ...
- 35.9k
11
votes
3
answers
561
views
Is the following Harmonic Number Identity true?
Is the following identity true?
$$ \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