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## A polyLogarithm value

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Moderator: Shobhit

### A polyLogarithm value

Thu Jul 11, 2013 6:34 pm
zaidalyafey Global Moderator

Posts: 357
I am looking for a way to evaluate $\displaystyle \text{Li}_3\left( \frac{1}{2}\right)$ ? , any hints ?

If there are some identities for trilogarithm I will be happy to know .
Wanna learn what we discuss , see Book

### Re: A polyLogarithm value

Thu Jul 11, 2013 7:14 pm
galactus
Global Moderator

Posts: 902
Any identities for trilogs would be in Lewin's book.

I suppose you already know that it is equal to:

$\displaystyle \frac{21}{24}\zeta(3)+\frac{1}{6}\ln^{3}(2)-\frac{{\pi}^{2}}{12}\ln(2)$

I found a few identities on wolframs site and in wikipedia.

But, I reckon you mean you want to evaluate $\displaystyle \sum_{k=1}^{\infty}\frac{1}{2^{k}k^{3}}$

### Re: A polyLogarithm value

Thu Jul 11, 2013 7:37 pm
zaidalyafey Global Moderator

Posts: 357
Ok, but is there an identity used for a general case

$\displaystyle \text{Li}_3(x) +\text{Li}_3(1-x)$
Wanna learn what we discuss , see Book

### Re: A polyLogarithm value

Thu Jul 11, 2013 8:19 pm
galactus
Global Moderator

Posts: 902
Hey Z:

Yeah, I think there is one.

Namely,

$\displaystyle \boxed{Li_{3}(x)+Li_{3}(1-x)+Li_{3}(1-\frac{1}{x})=\zeta(3)+\frac{\ln^{3}(x)}{6}+\frac{{\pi}^{2}\ln(x)}{6}-\frac{\ln^{2}(x)\ln(1-x)}{2}}$

This formula can be used to find $\displaystyle Li_{3}(1/2)$ because $\displaystyle Li_{3}(-1)=\frac{-3}{4}\zeta(3)$

### Re: A polyLogarithm value

Fri Jul 12, 2013 5:41 am
Shobhit Site Admin

Posts: 852
Location: Jaipur, India

Actually, there are thousands of Polylogarithm Identities ...

there are 8 functional equations for trilogarithm.

\displaystyle \begin{align*} \frac{1}{4}\text{Li}_3(x^2) &=\text{Li}_3(x)+\text{Li}_3(-x) \tag{1} \\ \text{Li}_3(-x)-\text{Li}_3(-1/x) &=-\frac{\pi^2}{6}\log x-\frac{1}{6}\log^3 x \tag{2}\\ \text{Li}_3(x)-\text{Li}_3(1/x) &=\frac{\pi^3}{3}\log(x)-\frac{1}{6} \log^3 x-\frac{1}{2}i \pi \log^2 x \tag{3}\\ \text{Li}_3 \left( \frac{-x}{1+x}\right)+\text{Li}_3 (1-x) +\text{Li}_3(x) &=\text{Li}_3(1)+\frac{\pi^2}{6} \log(1-x)-\frac{1}{2}\log(x) \log^2(1-x) \\ &\quad +\frac{1}{6}\log^3(1-x) \tag{4} \\ \text{Li}_3 \left( \frac{x}{1+x}\right)+\text{Li}_3 \left( \frac{1}{1+x}\right) +\text{Li}_3(-x) &=\text{Li}_3(1)+\frac{\pi^2}{6} \log(1+x)-\frac{1}{2}\log(x) \log^2(1+x) \\ &\quad +\frac{1}{3}\log^3(1+x) \tag{5} \\ \text{Li}_3 \left( 1-\frac{1}{x}\right)+\text{Li}_3(1-x)+\text{Li}_3(x) &= \text{Li}_3(1)+\frac{\pi^2}{6}\log(x)+\frac{1}{6}\log^3(x) \\ &\quad -\frac{1}{2}\log^2 x \log(1-x) \tag{6}\\ \text{Li}_3 \left( \frac{1-x}{1+x}\right)-\text{Li}_3 \left( \frac{x-1}{x+1}\right) &= 2\text{Li}_3(1-x)+2\text{Li}_3 \left( \frac{1}{1+x}\right)-\frac{1}{2} \text{Li}_3(1-x^2) - \frac{7}{4} \text{Li}_3(1) \\ &\quad +\frac{\pi^2}{6} \log(1+x) -\frac{1}{3}\log^3(1+x) \tag{7} \\ \end{align*}

By the way, you should read lewin's book. There are hundreds more...

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