Questions tagged [polylogarithm]

For questions about or related to polylogarithm functions.

The polylogarithm function $\operatorname{Li}_s$ is defined by the infinite sum

\begin{align*} \sum_{k = 1}^{\infty} \frac{z^k}{k^s} \end{align*}

for all $|z| < 1$ and complex order $s$, and obtained by analytic continuation of the sum. Depending on the order $s$, a branch cut must be taken for the logarithm.

In particular cases, the polylogarithm may have simpler representations; for example,

\begin{align*} \operatorname{Li}_1(z) &= -\ln{(1 - z)} \\ \operatorname{Li}_0(z) &= \frac{z}{1 - z} \\ \end{align*}

In the cases $s = 2$ and $s = 3$, the function is called the dilogarithm and trilogarithm, respectively.

The polylogarithm functions arise in quantum statistics and electrodynamics, and are related to the Fermi-Dirac integral.

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Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $

I'm looking for a closed form of this integral. $$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$ where $\operatorname{Li}_2$ is the dilogarithm function. A numerical approximation of it is $$ I \approx…
user153012
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Triple Euler sum result $\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$

In the following thread I arrived at the following result $$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$ Defining $$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$ But, it was after long evaluations and…
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Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$

Here is another integral I'm trying to evaluate: $$I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx.\tag1$$ A numeric approximation is: $$I\approx-0.19902842515384155925817158058508204141843184171999583129...\tag2$$ (click here to see more…
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Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$

$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ? Motivation: We already know that…
Lucian
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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}_4(\frac{1}{2}){}^2-\frac{19}{15} \pi ^4…
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A difficult logarithmic integral ${\Large\int}_0^1\log(x)\,\log(2+x)\,\log(1+x)\,\log\left(1+x^{-1}\right)dx$

A friend of mine shared this problem with me. As he was told, this integral can be evaluated in a closed form (the result may involve polylogarithms). Despite all our efforts, so far we have not achieved anything, so I decided to ask for your…
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Integral ${\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx$

This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here: Prove: …
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What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$?

Some time ago I asked How to find $\displaystyle{\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$. Thanks to great effort of several MSE users, we now know that \begin{align} \int_0^1\frac{\ln^3(1+x)\,\ln…
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A closed form for a lot of integrals on the logarithm

One problem that has been bugging me all this summer is as follows: a) Calculate $$I_3=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \ln{(1-x)} \ln{(1-xy)} \ln{(1-xyz)} \,\mathrm{d}x\, \mathrm{d}y\, \mathrm{d}z.$$ b) More generally, let $n \ge 1$ be an…
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Prove that $\int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3}$

I have spent my holiday on Sunday to crack several integral & series problems and I am having trouble to prove the following integral \begin{equation} \int_0^1 \frac{{\rm{Li}}_2(x)\ln(1-x)\ln^2(x)}{x} \,dx=-\frac{\zeta(6)}{3} \end{equation} Using…
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Conjectural closed-form of $\int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$

Let $$I_n = \int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$$ In a recently published article, $I_n$ are evaluated for $n\leq 6$: $$\begin{aligned}I_1 &= \frac{\log ^2(2)}{2}-\frac{\pi ^2}{12} \\ I_2 &= 2 \zeta (3) \log (2)-\frac{\pi…
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Yet another log-sin integral $\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\sin x)\,dx$

There has been much interest to various log-trig integrals on this site (e.g. see [1][2][3][4][5][6][7][8][9]). Here is another one I'm trying to solve: $$\int\limits_0^{\pi/3}\log(1+\sin x)\log(1-\sin x)\,dx\approx-0.41142425522824105371...$$ I…
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Integrals of integer powers of dilogarithm function

I'm interested in evaluating integrals of positive integer powers of the dilogarithm function. I'd like to see the general case tackled if possible, or barring that then as many particular cases as possible. Problem. For each $n\in\mathbb{N}$,…
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Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$

Please help me to evaluate this integral in a closed form: $$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$ Using integration by parts I found that it could be expressed through integrals of elementary…
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Show that $\sum_{n=0}^{\infty}\frac{2^n(5n^5+5n^4+5n^3+5n^2-9n+9)}{(2n+1)(2n+2)(2n+3){2n\choose n}}=\frac{9\pi^2}{8}$

I don't how prove this series and I have try look through maths world and Wikipedia on sum for help but no use at all, so please help me to prove this series. How to show…
user334593
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