Questions tagged [polylogarithm]

For questions about or related to polylogarithm functions.

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Generalised Polylogarithm Polynomials and Related Integer Sequences

Consider the generalised infinite summation $$S_{n,m}=m^{n+1} \sum_{k=1}^\infty \frac{k^n}{(m+1)^k}=m^{n+1}\,\mathrm{Li}_{(-n)} \left(\frac{1}{m+1}\right)$$ where $m$ and $n$ are positive integers, ...
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$\int\frac{\ln^kx\ln(1-x)}x dx$ Vs. $\int\frac{\ln x\ln^k(1-x)}x dx$

By using WolframAlpha, experimentely I observed that $$\frac1{k!}\int\frac{\ln^kx\ln(1-x)}x dx=(-1)^{k+1}\text{Li}_{k+2}(x)+\sum_{i=1}^k \frac{(-1)^{k+1-i}}{i!} \text{Li}_{k+2-i}(x)\ln^{i}x+c.$$ On ...
Bob Dobbs's user avatar
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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 ...
QuantumLogarithm's user avatar
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The ultimate polylogarithm ladder

As you can see, here I performed a derivation of a quite simple formula, not much differing from the standard integral representation of the Polylogarithm. Seeking to make it fancier, I arrived at ...
Artur Wiadrowski's user avatar
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3 answers
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Evaluating an integral from 0 to 1 with a parameter, (and a dilogarithm)

So I need to evaluate the following integral (in terms of a): $$\int_{0}^{1} \frac{\ln{|1-\frac{y}{a}|}}{y} dy$$ Till now I have tried u-sub ($u = \ln{|1-\frac{y}{a}|}$, $u=\frac{y}{a}$) and ...
CeaealYT's user avatar
11 votes
1 answer
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A cool integral: $\int^{\ln{\phi}}_{0}\ln\left(e^{x}-e^{-x}\right)dx=-\frac{\pi^2}{20}$

I was looking at the equation $\ln{e^{x}-e^{-x}}$ and found that the zero was at $x=\ln{\phi}$ where $\phi$ is the golden ratio. I thought that was pretty cool so I attempted to find the integral. I ...
Dylan Levine's user avatar
3 votes
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70 views

Show that $\mathfrak{Re}(\textrm{Li}_2(e^{ix}))=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$ (polylogarithm)

I am working with the polylogarithm function and want to find closed expressions for $\textrm{Li}_2(e^{ix})$. If I plot the function $\mathfrak{Re}(\textrm{Li}_2(e^{ix}))$ I get $y=\dfrac{x^2}{4}-\...
garondal's user avatar
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Solve the integral $\int_0^1 \frac{\ln^2(x+1)-\ln\left(\frac{2x}{x^2+1}\right)\ln x+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx$

I tried to solve this integral and got it, I showed firstly $$\int_0^1 \frac{\ln^2(x+1)+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=2\Im\left[\text{Li}_3(1+i) \right] $$ and for other integral $$\int_0^...
Faoler's user avatar
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Calculate an integral involving polylog functions

Im my recent answer https://math.stackexchange.com/a/4777055/198592 I found numerically that the following integral has a very simple result $$i = \int_0^1 \frac{\text{Li}_2\left(\frac{i\; t}{\sqrt{1-...
Dr. Wolfgang Hintze's user avatar
8 votes
1 answer
441 views

Prove $\int_{0}^{1}\frac1k K(k)\ln\left[\frac{\left(1+k \right)^3}{1-k} \right]\text{d}k=\frac{\pi^3}{4}$

Is it possible to show $$ \int_{0}^{1}\frac{K(k)\ln\left[\tfrac{\left ( 1+k \right)^3}{1-k} \right] }{k} \text{d}k=\frac{\pi^3}{4}\;\;? $$ where $K(k)$ is the complete elliptic integral of the first ...
Setness Ramesory's user avatar
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Closed form of dilogarithm fucntion involving many arctangents

I am trying to find closed form for this expression: $$ - 2{\text{L}}{{\text{i}}_2}\left( {\frac{1}{3}} \right) - {\text{L}}{{\text{i}}_2}\left( {\frac{1}{6}\left( {1 + i\sqrt 2 } \right)} \right) - {\...
OnTheWay's user avatar
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Polylogarithm further generalized

Here I proposed a generalized formula for the polylogarithm. However, because of a slight mistake towards the end, visible prior to the edit, I was unaware that it yields just a result of an integral ...
Artur Wiadrowski's user avatar
5 votes
1 answer
170 views

Evaluating $\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{(ax)}\operatorname{arsinh}{(bx)}}{x}$ in terms of polylogarithms

Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by the definite integral $$\mathcal{I}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arsinh}{\left(ax\right)}\...
David H's user avatar
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2 votes
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Complex polylogarithm/Clausen function/Fourier series

Sorry for the confusing title but I'm having a problem and I can phrase the question in multiple different ways. I was calculating with WolframAlpha $$\int \text{atanh}(\cos(x))\mathrm{d}x= i \text{Li}...
Math Attack's user avatar
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Converting polylogarithms to Dirichlet L functions

When trying to simplify polylogarithms evaluated at some root of unity, namely $\text{Li}_s(\omega)$ for $\omega=e^{2\pi i ~r/n}$, it is reasonable to convert it to Hurwitz zeta functions or Dirichlet ...
Po1ynomial's user avatar
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1 vote
1 answer
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Imaginary part of the dilogarithm of an imaginary number

I am wondering if I can simplify $${\rm Im} \left[ {\rm Li}_2(i x)\right] \ , $$ in terms of more elementary functions, when $x$ is real (in particular, I am interested in $0<x<1$). I checked ...
Rudyard's user avatar
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Evaluating $\int\frac{\log(x+a)}{x}\,dx$ in terms of dilogarithms

As per the title, I evaluated $$\int\frac{\log(x+a)}{x}\,dx$$ And wanted to make sure my solution is correct, and if not, where I went wrong in my process. Here is my work. $$\int\frac{\log(x+a)}{x}\,...
Person's user avatar
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1 vote
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Verification of the generalized polylogarithm formula

Here I posted a generalized formula for the polylogarithm I had discovered. However, for $x=\frac{1}{2}$, $z=\frac{1}{2}$, $p=1$ wolfram alpha yields a result different than what the double integral ...
Artur Wiadrowski's user avatar
6 votes
2 answers
305 views

How to show $\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G$

I am trying to prove that $$\int_0^1\frac{\operatorname{Li}_2\left(\frac{1+x^2}{2}\right)}{1+x^2}dx=\ln(2)G,$$ where $G$ is the Catalan constant and $\operatorname{Li}_2(x)$ is the dilogarithm ...
Ali Shadhar's user avatar
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Efficient calculation for Lerch Transcendent Expression

I've encountered: $$\Phi(z, s, \alpha) = \sum_{k=0}^\infty \frac { z^k} {(k+\alpha)^s}.$$ When trying to compute: $$\frac{1}{x}\sum_{p=0}^m \frac{2}{(2p-1)\ x^{2p-1}}\ s.t. x\in\mathbb{N} =\ ???$$ ...
user3108815's user avatar
3 votes
0 answers
133 views

Prove that $-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$

Prove that $$-\int_{0}^{1}\frac{{\rm Li}_2(-x(1-x))}{x}\ dx=\frac{2}{5}\zeta(3)$$ where ${\rm Li}_2(x)$ is the Poly Logarithm function and $\zeta(s)$ is the Riemann zeta function Let $$I=-\int_{0}^{1}...
Max's user avatar
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Evaluate $\sum\limits_{n=0}^\infty\operatorname W(e^{e^{an}})x^n$ with Lambert W function

$\def\W{\operatorname W} \def\Li{\operatorname{Li}} $ Interested by $\sum_\limits{n=1}^\infty\frac{\W(n^2)}{n^2}$, here is an example where Lagrange reversion applies to a Lambert W sum: $$\W(x)=\ln(...
Тyma Gaidash's user avatar
5 votes
1 answer
268 views

Closed forms of the integral $ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x $

(This is related to this question). How would one find the closed forms the integral $$ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x? $$ I tried using Nielsen Generalized Polylogarithm as mentioned ...
Anomaly's user avatar
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What's a better time complexity, $O(\log^2(n))$, or $O(n)$?

(By $O(\log^2(n))$, I mean $O((\log n)^2)$ rather than $O(\log(\log(n)) )$ I know $O(\log(n))$ is better than $O(\log^2(n))$ which itself is better than $O(\log^3(n))$ etc. But how do these compare to ...
Henry Deutsch's user avatar
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1 answer
31 views

Proving $2^{\log^{1-\varepsilon}n}\in \omega(\operatorname{polylog}(n))$

Let $\varepsilon \in (0,1)$. I wish to show that $2^{\log^{1-\varepsilon}(n)}\in \omega(\operatorname{polylog}(n))$. I attempted to turn this into a function and use L'Hospital's rule but that got me ...
Michal Dvořák's user avatar
1 vote
1 answer
119 views

Show that $\int_0^1 \frac{Li_{1 - 2m}(1 - 1/x)}{x} dx = 0$.

I would like to show that,for $m \geq 2$, $$I_m := \int_0^1 \frac{\operatorname{Li}_{1 - 2m}(1 - 1/x)}{x} dx = 0$$ where $\operatorname{Li}_{1 - 2m}$ is the $1-2m$ polylogarithm (https://en.wikipedia....
jvc's user avatar
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4 votes
0 answers
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How to derive this polylogarithm identity (involving Bernoulli polynomials)?

How can one derive the following identity, found here, relating the polylogarithm functions to Bernoulli polynomials? $$\operatorname{Li}_n(z)+(-1)^n\operatorname{Li}_n(1/z)=-\frac{(2\pi i)^n}{n!}B_n\!...
WillG's user avatar
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2 votes
1 answer
230 views

Generalized formula for the polylogarithm

Some time ago, I discovered the formula for repeated application of $z\frac{d}{dz}$ here. Recently, I thought about taking the function to which this would be applied to be the integral representation ...
Artur Wiadrowski's user avatar
5 votes
3 answers
248 views

How to find the exact value of $\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n^2 \cdot 2^{\frac{n}{2}}} $?

Once I met the identity $$ \boxed{S_0=\sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{2^{\frac{n}{2}}}=1}, $$ I first tried to prove it by $e^{xi}=\cos x+i\sin x$. $$ \begin{aligned} \...
Lai's user avatar
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2 votes
1 answer
82 views

Function upper-bounding polylogarithm function when $0 < z < 1$

The polylogarithm function (aka Jonquière's function) is defined as $Li_s(z) = \sum_{k=1}^{\infty} z^k k^{-s}$. Is there a closed-form upper bound for this function when $s \leq -1$ and is real, and $...
Daniel-耶稣活着's user avatar
5 votes
1 answer
255 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 ...
logandetner's user avatar
1 vote
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Series with power of generalized harmonic number $\displaystyle\sum_{k=1}^{\infty}\left(H_k^{(s)}\right)^n x^k$

It's possible to generalize these series? $$\sum_{k=1}^{\infty}H_k^{(s)}x^k=\frac{\operatorname{Li}_s(x)}{1-x}$$ $$\sum_{k=1}^{\infty}H_k^2 x^k=\frac{\ln(1-x)^2+\operatorname{Li}_2(x)}{1-x}$$ Where: $$...
Math Attack's user avatar
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1 vote
2 answers
84 views

Finding a recurrence relation to evaluate $\int_{a}^{1}\mathrm{d}x\,\frac{x^{n}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}$

For each $n\in\mathbb{Z}_{\ge0}$, define the function $\mathcal{J}_{n}:(0,1)\rightarrow\mathbb{R}$ via the doubly improper integral $$\mathcal{J}_{n}{\left(a\right)}:=\int_{a}^{1}\mathrm{d}x\,\frac{x^{...
David H's user avatar
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1 vote
2 answers
83 views

Asymptotics of an integral involving the exponential integral

Consider the integral: $$ I(a)=\int_a^\infty e^x E_1(x)\dfrac{dx}{x}, $$ where $a>0$ and $E_1(x)$ is the exponential integral function. I would like to better understand the behavior of $I(a)$ for $...
Jason's user avatar
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0 answers
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Solving a set of implicit equations involving Polylogarithms

I have the following simultaneous equations: \begin{aligned} &H(\lambda) = a\, \text{Li}_{3/2}\left(b\frac{H(\lambda)}{F(\lambda)}\right), \; \\&H(\lambda) = c\, \text{Li}_{3/2}\left(d \, \...
Harshit Rajgadia's user avatar
3 votes
2 answers
275 views

Imaginary part of dilogarithm

I have evaluated a certain real-valued, finite integral with no general elementary solution, but which I have been able to prove equals the imaginary part of some dilogarithms and can write in the ...
user47363's user avatar
0 votes
1 answer
74 views

Converting a 2D lattice sum into a sum over 1D lattice sums in a circle

I'm working on a physics problem. I have a lattice sum, which in 1D is a sum over a linear chain. It reads $$ f_k = \sum_{n=1}^\infty \frac{2\cos(kn)}{n^3}. $$ This can be written in terms of ...
Tom's user avatar
  • 501
2 votes
1 answer
169 views

Calculate the integral of the given polylogarithm function? $\int_0^1\frac{\operatorname{Li}_ 4(x)}{1+x}dx=?$ [closed]

$$\int_0^1 \frac{\operatorname{Li}_2(-x)\operatorname{Li}_2(x)}{x}\,\mathrm dx=?$$ where $$\operatorname{Li}_2(-x)=\sum_{k=1}^{\infty}\frac{(-x)^k}{k^2}$$ for $$|x|>1$$ actually my goal is to edit ...
merve kaya's user avatar
2 votes
1 answer
68 views

How to solve the indefinite integral $\int x \cot x\,\mathrm dx$

We can easily solve it with the bounds $0$ to $\frac{\pi}{2}$, but how to solve the indefinite integral? Wolfram Alpha gives the following solution: $$\int x \cot(x)\,\mathrm dx = x \log\left(1 - e^{2 ...
Bikram Kumar's user avatar
3 votes
1 answer
208 views

How to solve $\int\frac{x\arctan x}{x^4+1}dx$ in a practical way

I need to evaluate the following indefinite integral for some other definite integral $$\int\frac{x\arctan x}{x^4+1}dx$$ I found that $$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{...
phi-rate's user avatar
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21 votes
3 answers
2k views

Evaluate $\int_0^1\arcsin^2(\frac{\sqrt{-x}}{2}) (\log^3 x) (\frac{8}{1+x}+\frac{1}{x}) \, dx$

Here is an interesting integral, which is equivalent to the title $$\tag{1}\int_0^1 \log ^2\left(\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}\right) (\log ^3x) \left(\frac{8}{1+x}+\frac{1}{x}\right) \, dx =...
pisco's user avatar
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0 votes
1 answer
65 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^{...
Adolf L.'s user avatar
1 vote
0 answers
33 views

Dilogarithm Function on Negative Domain

I'm not that good with math, but somehow ended up solving for $ \int { \ln { (\cosh x) } } \cdot dx $. This has led me to the answer described here. In my case, I need a solution for x > 1, ...
Silver Flash's user avatar
3 votes
1 answer
139 views

Explicit value of $\operatorname{Li}_2(1/2-{\rm i}/2)$

When you ask Wolfram Alpha about the value of $\operatorname{Li}_{2}\left(1/2-{\rm i}/2\right)$ it gives you $$ \frac{5\pi^{2}}{96} - \frac{\ln^{2}\left(2\right)}{8} + {\rm i}\left[\frac{\pi\ln\left(2\...
Òscar Pérez Massanet's user avatar
3 votes
1 answer
140 views

I am stuck in this integral: $\int_{\sqrt{3}}^{\infty} \frac{\ln(x-1)}{x^2-1}dx$

I reduced one of the bounty problems to sum of some integrals. One of them is the following: $$\int_{\sqrt{3}}^{\infty} \frac{\ln(x-1)}{x^2-1}dx$$ It was a very complicated integral. And nobody gave a ...
Bob Dobbs's user avatar
  • 8,694
36 votes
8 answers
2k views

How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$

I am trying to find a closed form for $$ \int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right) {{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}. $$ I have done trig substitution and it results in $$ \int_{0}^{...
mike's user avatar
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1 vote
0 answers
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Can $\text{Li}_2(x/y)$ be expressed as a sum of other $\text{Li}_2$?

I am looking at the following function: $$f(x_1,x_2):=\text{Li}_2 \left( \frac{x_1}{x_2} \right). \tag{1}$$ I would like to know whether this function can be expressed as a sum of harmonic polylogs (...
Pxx's user avatar
  • 697
0 votes
2 answers
109 views

Polylog integral $\int_{0}^{1}\frac{x\log x\operatorname{Li}_2(x)}{x^2+1}dx$

I am trying to solve this: $\newcommand{\dilog}{\operatorname{Li}_2}$ $$\int_{0}^{1}\frac{x\log x\dilog(x)}{x^2+1}dx$$ My try: use the definition of $\dilog(x)$ as $$\dilog(x)=\sum_{k=1}^{\infty}\frac{...
OnTheWay's user avatar
  • 2,288
1 vote
1 answer
53 views

Bounding fractional moments of geometric random variable

The following two bounds for a fractional moment of a geometric random variable $X$ with $\mathbb{P}\left[X = k\right] = p \left(1 - p\right)^k$ where $k \geq 0$ are given in this paper (on page 12): ...
M_F's user avatar
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2 votes
0 answers
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The summation $\sum_{k=1}^{\infty} \frac{\phi^{-2k}}{k^2}=\text{Li}_2(1/\phi^2)=\frac{\pi^2}{15}-\ln^2 \phi$ [duplicate]

In a recent post Computing $\int_0^{1/2}\frac{\sinh^{-1}(u)}{u} \,du=\frac{\pi^2}{20}$, $\zeta(2)=\frac53 \sum_{n=0}^\infty \frac{(-1)^n\binom{2n}{n}}{2^{4n}(2n+1)^2}$ I evaluated the required ...
Z Ahmed's user avatar
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