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Let \begin{align} Li_2(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}. \end{align}

There are two different forms of Abel identities for polylogarithms:

1. \begin{align} & Li_2(-x) + \log x \log y \\ & + Li_2(-y) + \log ( \frac{1+y}{x} ) \log y \\ & + Li_2(-\frac{1+y}{x}) + \log ( \frac{1+y}{x} ) \log (\frac{1+x+y}{xy}) \\ & + Li_2(-\frac{1+x+y}{xy}) + \log ( \frac{1+x}{y} ) \log (\frac{1+x+y}{xy}) \\ & + Li_2(-\frac{1+x}{y}) + \log ( \frac{1+x}{y} ) \log x \\ & = - \frac{\pi^2}{2}. \end{align} 2. \begin{align} & Li_2(x) + Li_2(y) - Li_2(xy) \\ & = Li_2(\frac{x(1-y)}{1-xy}) + Li_2(\frac{y(1-x)}{1-xy}) + \log(\frac{1-x}{1-xy}) \log(\frac{1-y}{1-xy}). \end{align} Are there some references of proving that these two identities are equivalent? Thank you very much.

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    I think you have to explain what is meant with *equivalent* here! E.g. using derivation (so that we only have simple expressions) the two identities *don't* come together. – 2017-03-01
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    @user90369, these formulas appears in [the thesis](https://www.brown.edu/academics/physics/sites/brown.edu.academics.physics/files/uploads/Parker-2015.pdf) (6.10), (6.63). It is said that they are two different forms of Abel identities. So I think that they are equivalent. – 2017-03-01

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My comment is a bit too long therefore I write it as an answer.

Thanks for the link, it makes the question much clearer. With $(6.63)$ follows directly $2.$ by the substitution $x\to\frac{x}{1-y}$ and $y\to\frac{y}{1-x}$ . The script uses the symbol-technique $S$ to transform the terms of $(6.63)$ into other terms getting an equation which is somehow equivalent to $1.$ . This equation now has to be retransformed to $1.$ and the constant must be still determined. So it depends on the definition of "equivalent" if $1.$ und $2.$ are equivalent. It's the same question, if e.g. $\frac{1}{(1-x)^2}=\sum\limits_{n=1}^\infty nx^{n-1}$ and $\frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n$ are equivalent. I would say: Because of the difference with the constant it's an incomplete Yes (that $1.$ and $2.$ are equivalent). I hope you will find a better expression than me. :-)