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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