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 .

Board index **‹** Special Functions **‹** A polyLogarithm value
## A polyLogarithm value

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 \(\displaystyle \sum_{k=1}^{\infty}\frac{1}{2^{k}k^{3}}\)

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 \(\displaystyle Li_{3}(-1)=\frac{-3}{4}\zeta(3)\)

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...

**Moderator:** Shobhit

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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 .

If there are some identities for trilogarithm I will be happy to know .

Wanna learn what we discuss , see 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 \(\displaystyle \sum_{k=1}^{\infty}\frac{1}{2^{k}k^{3}}\)

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

\(\displaystyle \text{Li}_3(x) +\text{Li}_3(1-x)\)

\(\displaystyle \text{Li}_3(x) +\text{Li}_3(1-x)\)

Wanna learn what we discuss , see Book

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 \(\displaystyle Li_{3}(-1)=\frac{-3}{4}\zeta(3)\)

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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