I would like to know if there is a proof for the identity used in the superconformal index of 4d ${\cal N}=2$ gauge theory. In the [paper](http://arxiv.org/pdf/1104.3850v1.pdf) by Rastelli el al, it was discovered that Eq. (10) is equal to the right hand side in the previous equation. 
\begin{equation}
\exp\left[\sum\_{n>0}\frac{1}{n}\frac{q^{n/2}}{1-q^n}\chi\_{[1]}(a\_1^n)\chi\_{[1]}(a\_2^n)\chi\_{[1]}(a\_3^n) \right]=\frac{(q)\_\infty}{1-q}\prod^3\_i \eta\_2^{-1/2}(a\_i) \sum\_{R}\frac{\chi\_R(a\_1)\chi\_R(a\_2)\chi\_R(a\_3)}{\dim\_q R}
\end{equation}
wheret
\begin{equation}
\eta\_2(x)=\exp\left[-2\sum\_{n>0} \frac{1}{n}\frac{q^n}{1-q}(\chi\_{[1]}^2(x)-1)\right]
\end{equation}
Note that $\chi\_R$ is a character of the irreducible representation of ${\mathfrak sl}\_2$ with highest weight $R$. If there are only two characters involved, it could be seen as the Cauchy formula in representation theory. I wonder if somebody know a proof for this identity.



Proof of generalized Cauchy formula
1) In appendix E of [this paper](http://arxiv.org/pdf/1110.3740.pdf) there is an outline of a proof of this statement based on matching poles and residues on both sides of the identity. 


2) One can also use the more generic arguments of [a more recent paper](http://arxiv.org/pdf/1207.3577.pdf) even for the more  

general identity involving Macdonald polynomials. One can show that acting on
the left hand side with Macdonald operator (properly conjugated) on any of the $a\_i$ 
gives the same result independent on the choice of $a\_i$. see eg 5.11 there. 
then one can expand the left hand side in terms of the Macdonald polynomials which
is given by a single sum over representations since the spectrum of Macdonald operator 
is not degenerate. Using this observation and some simple manipulations the identity can be established - the complete argument is detailed in [section 6](http://arxiv.org/pdf/1207.3577.pdf) of the above mentioned paper.


