Questions tagged [polylogarithm]
For questions about or related to polylogarithm functions.
546
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General expression of a (maybe 3 or 2 dim) sequence [closed]
$\frac{1}{2}$
$\frac{1}{4}$ $\frac{1}{2}$
$\frac{1}{6}$ $\frac{1}{4}$ $\frac{11}{24}$
$\frac{1}{8}$ $\frac{1}{6}$ $\frac{11}{48}$ $\frac{5}{12}$
$\frac{1}{10}$ $\frac{1}{8}$ $\frac{11}{...
- 99
5
votes
1
answer
223
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How to evaluate $\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$
I want to evaluate $$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\right)\:dx$$
But I've not been successful in doing so, what I tried is
$$\int _0^1\ln \left(\operatorname{Li}_2\left(x\right)\...
user809806
2
votes
2
answers
288
views
Does the dilogarithm function (which is multi-valued) have a single-valued inverse?
The $p$-logarithm is defined for $|z|<1$ by
$$\text{Li}_p(z)=\sum_{n=1}^\infty\frac{z^n}{n^p}$$
and defined elsewhere in $\mathbb C$ by analytic continuation, though it may be multi-valued, ...
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0
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Further question on Logarithm product integral
How to perform $\int_0^1 \frac{\left(a_0\log(u)+a_1\log(1-u)+a_{2}\log(1-xu)\right)^9}{u-1} du $?
Method tried:
Intgration-by-parts
Series expansion
change of variable $\log(u)=x$
But I still can't ...
- 99
2
votes
1
answer
138
views
Powers of polylogarithms
I would like to take powers of arbitrary order to polylogarithm functions. For instance, given
$$
\text{Li}_\alpha(z) = \sum_{k=1}^\infty \frac{z^k}{k^\alpha}
$$
I am interested in
$$
[\text{Li}_\...
- 551
17
votes
2
answers
894
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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
1
vote
0
answers
157
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Upper bounds regarding polylogarithm $Li_p(e^w)$ when $|w| > 2\pi$ and p negative real
$\textit{The Computations of Polylogarithms, 1992,Technical report UKC, University of Kent, Canterbury, UK}\\ $ by $\textbf{David C Wood}$ says that
$$
Li_p(e^w) = \sum_{n\geq 0} \zeta(p-n)\frac{w^n}{...
- 417
6
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Integral from Mathematica's documentation: $\int_0^1 \frac{\log (\frac{1}{2}(1+\sqrt{4 x+1}))}{x} \, dx = \frac{\pi^2}{15} $
I like to peruse Mathematica's documentation and look at the 'Neat Examples': this is one I managed to figure out. Apparently it's due to Ramanujan:
$$
I=\int_0^1 \frac{\log \left(\frac{1}{2} \left(1+\...
- 8,369
23
votes
3
answers
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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:
$$\int_0^{2\pi}x^2 \cos(x)\operatorname{Li}_2(\cos(x))dx$$
$$=\frac{9}{8}\pi^4-2\pi^3-2\pi^2-8\ln(2)\pi-\frac12\ln^2(2)\pi^2+8\ln(2)\pi G+16\pi\Im\left\{\...
- 25.8k
8
votes
1
answer
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How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$.
How can i evaluate
$$\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
1
answer
326
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Evaluate $\Im(\operatorname{Li}_3(2i) + \operatorname{Li}_3(\frac i2))$
Applying the trilogarithm identity
$$ \operatorname{Li}_{3}\left(z\right) - \operatorname{Li}_{3}\left(1 \over z\right) =
-{1 \over 6}\ln^{3}\left(-z\right) -
{\pi^{2} \over 6}\ln\left(-z\right)\tag{1}...
- 99.7k
2
votes
4
answers
157
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Errors are decreasing in series $\sum_{n=1}^\infty(-1)^n/n^4$?
Let $v=\sum_{n=1}^\infty(-1)^n/n^4$ ($v$ for "value"), let $S=(\sum_{n=1}^m(-1)^n/n^4)_{m\in\mathbb Z_{\ge1}}$ be the partial sums, and let $e=(|S_n-v|)_{n\in\mathbb Z_{\ge1}}$ be the errors....
- 867
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An integral involving a Gaussian and a power of a normal cumulative distribution function
Being inspired by How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$? we formulated the question below.
Let $c \in (0,1/\sqrt{2})$ and let $n \in \Bbb{N}$. Then let $\phi(x) : =\frac{\...
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Does there exist a closed form for $\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$?
I am not sure if there exists a closed form for
$$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$
which seems non-trivial.
I used the reflection and landen's identity, didn't help much.
...
- 25.8k
4
votes
2
answers
335
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How to approach $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$ elegantly?
How to show that
$$\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)$$
without using the generating function:
\...
- 25.8k
6
votes
1
answer
765
views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
- 25.8k
17
votes
2
answers
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How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\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
8
votes
4
answers
688
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How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$
Before you think I haven't tried anything, please read.
I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
But I can't find a way to simplify it. ...
user809806
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
5
votes
0
answers
281
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Is there a closed form without MZV for $ \sum _{k=1}^{\infty }\frac{H_k}{k^6\:2^k}$?
While evaluating the weight $7$ integral $\displaystyle \int_0^1\frac{\ln^3\left(1-x\right)\ln^3\left(1+x\right)}{1+x}\:dx$
I managed to prove that
$$\int_0^1\frac{\ln^3\left(1-x\right)\ln^3\left(1+x\...
- 2,646
9
votes
2
answers
941
views
Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$
Here is a nice problem proposed by Cornel Valean
$$
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
4
votes
2
answers
411
views
How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$
I want to evaluate
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$
Im not sure if this has a closed form, integration by parts is out of the question since there would be ...
user805212
6
votes
1
answer
414
views
How can I evaluate $\int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
I am trying to evaluate $\displaystyle \int _0^1\frac{\text{Li}_2\left(-x\right)\ln \left(1-x\right)}{1+x}\:dx$
I first tried using the series expansion for the dilogarithm like this
$$\sum _{n=1}^{\...
user806882
3
votes
1
answer
96
views
On $\int_0^{2\pi }\frac{\prod_{k=1}^m \text{Li}_{a_k}(e^{-ix})-\prod_{k=1}^m \text{Li}_{a_k}(e^{ix})}{e^{-ix}-e^{ix}} \, dx$
OP of this post evaluated a lot of remarkable polylog integrals (without proof), from which I conjecture a generalized one ($a_k\in \mathbb N$):
$$\int_0^{2 \pi } \frac{\prod _{k=1}^m \text{Li}_{a_k}(...
- 8,654
44
votes
2
answers
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Remarkable logarithmic integral $\int_0^1 \frac{\log^2 (1-x) \log^2 x \log^3(1+x)}{x}dx$
We have the following result ($\text{Li}_{n}$ being the polylogarithm):
$$\tag{*}\small{ \int_0^1 \log^2 (1-x) \log^2 x \log^3(1+x) \frac{dx}{x} = -168 \text{Li}_5(\frac{1}{2}) \zeta (3)+96 \text{Li}...
- 19.1k
1
vote
2
answers
83
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Closed-form expressions for the zeros of $\text{Li}_{-n}(x)$?
Consider the first few polylogarithm functions $\text{Li}_{-n}(x)$, where $-n$ is a negative integer and $x\in\mathbb R$ (plotted below). Observation suggests that $\text{Li}_{-1}(x)$ has one zero (at ...
- 6,682
2
votes
2
answers
226
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Evaluation of a log-trig integral in terms of the Clausen function (or other functions related to the dilogarithm)
Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(a,\theta\right)}:=\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2a\cos{\left(\...
- 30.7k
0
votes
1
answer
101
views
What's this partial sum $ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$ equal?
I want to get this partial sum of $$ \sum_{k=0}^{n-1} \dfrac{\log(k!)}{2^{k+1}}$$ which it is convergent and it is closed to one half , I have tried to use polylogarithm function which is defined as :...
- 7,195
5
votes
1
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429
views
Evaluate $\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$
I encountered a hypergeometric integral while investigating harmonic sums
$$\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$$
Based on my experience I suspect a nice ...
- 8,654
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
5
votes
3
answers
320
views
Is there a closed-form for $\sum_{n=0}^{\infty}\frac{n}{n^3+1}$?
I'm reading a book on complex variables (The Theory of Functions of a Complex Variable, Thorn 1953) and the following is shown:
Let $f(z)$ be holomorphic and single valued in $\mathbb{C}$ except at a ...
- 8,369
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.$$
...
user753116
3
votes
2
answers
187
views
Closed form for $\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x$
I am looking for a closed form for:
$$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$
I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/...
user753116
1
vote
1
answer
256
views
Looking for closed form for $\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x$
I was evaluating and integral involving iterated logarithms when the following integral appeared:
$$\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
0
votes
0
answers
34
views
How does $\lim\limits_{Re(s)\rightarrow -\infty} Li_s(e^{w})$ become $\Gamma(1-s)(-w)^{s-1}$?
I am trying to prove the following property of polylogarithm. $$\lim\limits_{Re(s)\rightarrow -\infty} Li_s(e^{w}) = \Gamma(1-s)(-w)^{s-1} \text{ for } -\pi < \Im(w) < \pi.$$
As per Wikipedia's "...
- 417
4
votes
3
answers
436
views
A closed form for the dilogarithm integral $\int _{ 0 }^{ 1 }{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx } $
$$\int _{ 0 }^{ 1 }{ \frac { \operatorname{Li}_2\left( 2x\left( 1-x \right) \right) }{ x } dx } $$
when I was solving an infinite series by using the beta function I encountered the above ...
- 247
2
votes
0
answers
49
views
Where am I wrong in my calculations involving the Polylogarithm?
I was messing around with some formulas for the Polylogarithm $\operatorname{Li}_s(z)$ when I got the result that $0 = \frac{\sqrt{\pi}}{2}$ which is clearly absurd. Here are my calculations:
Under ...
- 1,054
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:
$$10\...
user753116
2
votes
2
answers
241
views
Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$
How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$
Any help will be appreciated.
- 8,654
1
vote
0
answers
145
views
Dilogarithm of a negative real number outside unit circle
The dilogarithm is defined in $\mathbb{C}$ as
$$
Li_2(z) = -\int_0^1 \frac{\ln(1 - zt)}{t} dt
$$
Because $1-zt \in \mathbb{C}$, then you can write $\ln(1 - zt) = \ln|1 - zt| + i·\arg(1 - zt)$
As ...
- 539
2
votes
0
answers
154
views
Evaluating a variant of the polylogarithm
Consider the infinite sum :
$$\sum_{m=1}^{\infty}\binom{m}{my}\frac{z^{m}}{m^{s}}\;\;\;\;s\in\mathbb{C},\;\;\;\;|z|<\frac{1}{2},\;\;\;\;\;0<y<1$$
I want to evaluate this summation in terms ...
4
votes
1
answer
258
views
Could this integral expression for $\zeta(3)$ be simplified any further?
I managed to derive the following integral:
$$\zeta \left( s \right) ={\frac { \left( s-2 \right)}{\Gamma \left( s \right) } \int_{0}^{\infty }\!{u}^{s-3} \left( \zeta(2)-{\it Li_2} \left(1-{{\rm e}^{...
- 3,191
3
votes
4
answers
300
views
Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$
Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted ...
- 25.8k
0
votes
1
answer
97
views
dilogarithm property.
Prove that
$\mathrm{Li}_{2}(-z)+\mathrm{Li}_{2}\left(\frac{z}{1+z}\right)=-\frac{1}{2} \ln ^{2}(1+z)$
I tried to paint in the rows, but I did not succeed. I don't have any more ideas.
- 222
5
votes
1
answer
376
views
Evaluate $\int_0^1 \frac{x \operatorname{Li}_2(x) \log (1+x)}{x^2+1} \, dx$
$$\int_0^1 \frac{x \operatorname{Li}_2(x) \log (1+x)}{x^2+1} \, dx=-\frac{3\pi }{4} \Im(\operatorname{Li}_3(1+i))+\frac{189}{128} \zeta (3) \log (2)+\frac{C^2}{2}-\frac{1}{4} \pi C \log (2)+\frac{ \...
- 5,507