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$1)$ Write $\text{Li}_2(1-\frac{1}{x})$ in terms of $\text{Li}_2(x)$ and logarithms by considering its integral representation and suitable changes of variables.

Attempt: The di-log is defined as $$\text{Li}_2(z) = \int_0^z \frac{dt}{t} \text{Li}_{1}(t) = -\int_0^z \frac{dt}{t} \text{log}(1-t) .$$ I'm just trying to see how to start, I could set $z=1-1/x$ as a change of variables but I am not sure if this helps.

$2)$ Use the shuffle identities to expand the products

$A. G(a_1, a_2; z) G(a_3; z) $

Attempt: The shuffle product is a linear combination of polylogs of weight $2+1$ in this case which preserves the relative ordering of the indices $a_i$ in each of the terms. So if I understand this correctly, here we simply have $G(a_1, a_2,a_3;z) + G(a_1, a_3, a_2;z) + G(a_3, a_1,a_2;z)$ i.e the only restriciton is that $a_1$ is always to the left of $a_2$, i.e preserving the order. Is that right?

CAF
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1 Answers1

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We just need to prove the identities $(4)$ and $(5)$ stated here: $$ \text{Li}_2(1-z)+\text{Li}_2\left(1-\frac{1}{z}\right) = -\frac{1}{2}\log^2 z,\tag{4}$$ $$ \text{Li}_2(z)+\text{Li}_2(1-z)=\frac{\pi^2}{6}-\log z\, \log(1-z)\tag{5}$$ $(5)$ has already been proved in this question.

To prove $(4)$ it is enough to notice that $\text{Li}_2(0)=0$ and: $$ \frac{\log z}{1-z}+\frac{\log z}{z^2-z}=-\frac{\log z}{z},$$ so the values at $z=1$ and the derivatives of the LHS and RHS of $(4)$ match.

By combining $(4)$ and $(5)$ we get:

$$\text{Li}_2(z)-\text{Li}_2\left(1-\frac{1}{z}\right)=\frac{\pi^2}{6}-\log z\,\log(1-z)+\frac{1}{2}\log^2 z.$$

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Jack D'Aurizio
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  • Many thanks! Do you have any comments regarding whether my understanding of the shuffle algebra is correct? – CAF May 26 '15 at 21:17
  • I also looked at the proof by the other responder on the other thread. How did he get from the first line to the second line? In particular, why did the limits on the integral change the way they did? Thanks! – CAF May 26 '15 at 21:22
  • @CAF: I am not so confident in shuffle algebras but I would bet that the geometric invariants lead to objects that behave like dilogarithms, by your considerations. About the other question, he simply broke the integration range in two pieces and did a change of variable in the second integral. – Jack D'Aurizio May 26 '15 at 21:25