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A Trilogarithm Identity

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Post Sat Jan 25, 2014 5:05 am
Shobhit Site Admin
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\(\displaystyle \text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)=-\frac{\log^3 3}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}\)

Post Tue Nov 24, 2015 1:26 am

Posts: 48

Shobhit wrote:
Show that

\(\displaystyle \text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)=-\frac{\log^3 3}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}\)


Here is a solution.

We are using the identity:

$${\rm Li}_3 \left ( \frac{1-z}{1+z} \right ) - {\rm Li}_3 \left ( - \frac{1-z}{1+z} \right )= -2 {\rm Li}_3 \left ( \frac{z}{z-1} \right ) - 2{\rm Li}_3 \left ( \frac{z}{z+1} \right )+ \frac{1}{2}{\rm Li}_3 \left ( \frac{z^2}{z^2-1} \right )+ \frac{7}{4}{\rm Li}_3 (1)+ \\
+\frac{1}{4}\log \left ( \frac{1-z^2}{z^2} \right )\log^2 \left ( \frac{1+z}{1-z} \right )+ \frac{1}{4}\pi^2 \log \left ( \frac{1-z}{1+z} \right )$$

along with some known results of the trilog , i.e ${\rm Li}_3(1)=\zeta(3), \; {\rm Li}_3(-1)=-\frac{3\zeta(3)}{4}$. Hence subbing $z=1/2$ we have the desired result.

Could someone please tell me how they came up with the identity? :?


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