All Questions
Tagged with polylogarithm real-analysis
60
questions
1
vote
1
answer
112
views
asymptotic behaviour of polylogarithmic function
I would like to understand the asymptotic behaviour as $a \rightarrow 0$ of the function
$$
f(a) := \sum\limits_{k=2}^{\infty} e^{ - a^2 k}{k^{-3/2}}
$$
More precisely, I would like to obtain an ...
- 200
6
votes
1
answer
285
views
Calculate $\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$
this integral got posted on a mathematics group by a friend
$$I=\int _0^1\frac{\arcsin ^2\left(x\right)\ln \left(x\right)\ln \left(1-x\right)}{x}\:\mathrm{d}x$$
I tried seeing what I'd get from ...
- 231
0
votes
1
answer
93
views
$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$ as polylogarithm
$$\sum_{i=1}^{n} \frac {x^{2i-1}}{\sqrt{2i}}$$
It is very clear for me that it has to be polylogarithm function but as it is partial sum I tried to split the series as
$$\sum_{i=1}^{\infty} \frac {x^{...
- 3
3
votes
0
answers
316
views
Two tough integrals with logarithms and polylogarithms
The following two integrals are given in (Almost) Impossible Integrals, Sums, and Series (see Sect. $\textbf{1.55}$, page $35$),
$$i) \int_0^{\pi/2} \cot (x) \log (\cos (x)) \log ^2(\sin (x)) \...
- 5,495
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:
$$\int_0^1\...
- 25.8k
0
votes
1
answer
148
views
Evaluate: ${{\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx}}$
Evaluate:
$${{I=\int_{0}^{1}\frac{\ln(1+x)^5}{x+2}dx-\int_{0}^{1}\frac{\ln(1+x)^5}{x+3}dx+5\ln2\int_{0}^{1}\frac{\ln(1+x)^4}{x+3}dx.}}$$
The answer is given below:
$$
I=-\frac{7}{12}\pi^4\ln^2(2)-\...
- 5,370
1
vote
1
answer
164
views
Integral involving product of dilogarithm and an exponential
I am interested in the integral
\begin{equation}
\int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast)
\end{equation}
where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to ...
- 753
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
$$\mathcal{L}(2)=\int_{0}^{\...
- 5,370
10
votes
1
answer
790
views
A generalized "Rare" integral involving $\operatorname{Li}_3$
In my previous post, it can be shown that
$$\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
1
answer
1k
views
A rare integral involving $\operatorname{Li}_2$
A rare but interesting integral problem:
$$\int_{0}^{1}
\frac{\operatorname{Li}_2(-x)-
\operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x)
-\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} }
=\...
- 5,370
6
votes
1
answer
494
views
Is the closed form of $\int_0^1 \frac{x\ln^a(1+x)}{1+x^2}dx$ known in the literature?
We know how hard these integrals
$$\int_0^1 \frac{x\ln(1+x)}{1+x^2}dx;
\int_0^1 \frac{x\ln^2(1+x)}{1+x^2}dx;
\int_0^1 \frac{x\ln^3(1+x)}{1+x^2}dx;
...$$
can be. So I decided to come up with a ...
- 25.8k
3
votes
1
answer
146
views
Integral representation of Dilogarithm
I have a question regarding how to obtain a certain integral representation of the dilogarithm, namely:
$$\DeclareMathOperator{\li}{Li_2}\li(x) = -\int_{0}^{1}\frac{x\ln t}{1-tx}\mathrm dt\quad ,|x|&...
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
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:
$$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$
$$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\...
- 25.8k
8
votes
1
answer
497
views
How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$.
How can i evaluate
$$\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$$
$$=\frac{\pi ^3}{6}-\frac{\pi ^3}{6\sqrt{2}}-4\pi +6\pi \ln \left(2\right)-\frac{\pi }...
user809806
2
votes
4
answers
157
views
Errors are decreasing in series $\sum_{n=1}^\infty(-1)^n/n^4$?
Let $v=\sum_{n=1}^\infty(-1)^n/n^4$ ($v$ for "value"), let $S=(\sum_{n=1}^m(-1)^n/n^4)_{m\in\mathbb Z_{\ge1}}$ be the partial sums, and let $e=(|S_n-v|)_{n\in\mathbb Z_{\ge1}}$ be the errors....
- 867
6
votes
1
answer
765
views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
- 25.8k
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
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
- 25.8k
10
votes
4
answers
619
views
How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way?
The following problem
\begin{align}
&\int_{0}^{\pi/2}
x\ln^{2}\left(\sin\left(x\right)\right)\,{\rm d}x \\[5mm] = & \
\frac{1}{2}\ln^{2}\left(2\right)\zeta\left(2\right)
- \frac{19}{32}\,\zeta\...
- 25.8k
11
votes
2
answers
728
views
How to compute $\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx$ or $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$
How to tackle
$$I=\int_0^1\frac{\text{Li}_2(x^2)\arcsin^2(x)}{x}dx\ ?$$
This integral came up while I was working on finding $\sum_{n=1}^\infty\frac{4^nH_n}{n^4{2n\choose n}}$.
First attempt: By ...
- 25.8k
13
votes
3
answers
689
views
How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$
Here is a new challenging problem:
Show that
$$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\}$$...
- 25.8k
9
votes
2
answers
941
views
Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$
Here is a nice problem proposed by Cornel Valean
$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{...
- 25.8k
1
vote
1
answer
111
views
How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?
where $H_n$ denotes the harmonic number.
I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even
$$-\sum_{n\geq 1}H_{2n}^2\...
- 125
2
votes
1
answer
86
views
Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?
Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original ...
- 125
2
votes
1
answer
111
views
Evaluate $\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}$.
I am looking for a closed for $$\sum_{n\geq1} \frac{(-1)^{n+1}H_n^2}{(n+1)^2}.$$ I believe there is a closed form for the sum as we have seen in [1] which poses as, presumably, a more difficult sum of ...
- 125
1
vote
2
answers
150
views
How to evaluate $ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}$
I am having a difficult time evaulating
$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}.$$
I have tried the following relation:
$$\frac{1}{2}\int_0^1 \frac{\mathrm{Li}_2(x)}{x(1-x)}\log^2{x}\ \mathrm{d}x.$$
...
user753116
1
vote
1
answer
256
views
Looking for closed form for $\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x$
I was evaluating and integral involving iterated logarithms when the following integral appeared:
$$\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \...
user753116
8
votes
2
answers
270
views
Is there a closed form for $\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x$?
Do we know if there is a closed form for
$$
I :=\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x\mathrm{?}
$$
Wolfram alpha gives an approximation of $2.66989$ which may be equivalent to:
$$10\...
user753116
5
votes
2
answers
452
views
Calculating $\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx$
How to prove in a simpe way that
$$\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx=\frac{13}{8}\ln2\zeta(2)-\frac{33}{32}\zeta(3)\ ?$$
where $\operatorname{Li}_2$ is the ...
- 25.8k
2
votes
1
answer
291
views
How to compute $\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\ln^2(1-x^2)}{1-x^2}\ dx$
How to compute
$$\int_0^1\left(\operatorname{Li}_2(x)-\zeta(2)\right)\frac{\ln^2(1-x^2)}{1-x^2}\ dx$$
where $\operatorname{Li}_2(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$ is the dilogarithm function.
...
- 25.8k
4
votes
1
answer
439
views
Advanced Sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}$
How to prove
$$\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?
$$ \sum_{n=1}^\infty \frac{H_nx^n}{n^3} = \frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(...
- 5,558
16
votes
2
answers
889
views
Evaluate $ \int_{0}^{1} \log\left(\frac{x^2-2x-4}{x^2+2x-4}\right) \frac{\mathrm{d}x}{\sqrt{1-x^2}} $
Evaluate :
$$ \int_{0}^{1} \log\left(\dfrac{x^2-2x-4}{x^2+2x-4}\right) \dfrac{\mathrm{d}x}{\sqrt{1-x^2}} $$
Introduction : I have a friend on another math platform who proposed a summation ...
- 5,558
3
votes
1
answer
513
views
Another way of doing integration
What's your option for calculating this integral? No full solution is necessary, it's optional as usual.
Calculate
$$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \...
- 44.4k
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
6
votes
2
answers
201
views
Double integral with a product of dilog $\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x)\ dx \ dy$
One of the integrals I came across these days (during my studies) is $$\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x) \ dx \ dy$$
that can be turned into a series, or can be approached by ...
- 44.4k
4
votes
1
answer
330
views
A possible dilogarithm identity?
I'm curious to find out if the sum can be expressed in some known constants. What do you think about that? Is it feasible? Have you met it before?
$$2 \left(\text{Li}_2\left(2-\sqrt{2}\right)+\text{...
- 44.4k