Questions tagged [polylogarithm]

For questions about or related to polylogarithm functions.

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

$$\operatorname{Li}_s(z)=\sum_{k = 1}^{\infty} \frac{z^k}{k^s}$$

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} \\ ...\ &=\ ...\\ \operatorname{Li}_{n-1}(z)&=z\frac{\mathrm{d}}{\mathrm{d}z}\text{Li}_n(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.

Is directly related to the function: $$\text{Li}_s(1)=\zeta(s)\qquad \text{for }s>1$$

Furthermore, its derivatives with respect to the parameter $s$ are also defined:

$$\text{Li}_s^{(n,0)}(z):=\frac{\partial^n}{\partial s^{s}}\text{Li}_s(z)=\sum_{k=1}^{\infty}(-1)^n\sum_{k=1}^{\infty}\frac{\ln(k)^n}{k^s}z^k\qquad\text{for }|z|<1$$

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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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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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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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Are these generalizations known in the literature?

By using $$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$ and $$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(2k)}{(a-2k)!}\ln^{a-2k}(z)\tag{b}$$ I managed to…
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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}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right) {{\rm d}x…
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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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On the relationship between $\Re\operatorname{Li}_n(1+i)$ and $\operatorname{Li}_n(1/2)$ when $n\ge5$

Motivation $\newcommand{Li}{\operatorname{Li}}$ It is already known that: $$\Re\Li_2(1+i)=\frac{\pi^2}{16}$$ $$\Re\Li_3(1+i)=\frac{\pi^2\ln2}{32}+\frac{35}{64}\zeta(3)$$ And by this question, assuming $\Li(1/2)$ is a simpler form than $\Re\Li(1+i)$,…
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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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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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Evaluate $\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right)\:\mathrm{d}x$

How can I prove in an elementary way that: $$\int _0^{\pi /2}x\operatorname{Li}_3\left(\sin \left(x\right)\right)\:\mathrm{d}x=\int _0^1\frac{\arcsin…
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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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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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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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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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