Itzykson-Zuber integral is defined as
We are free to assume that both and are diagonal, since if they are not (though they should be Hermitian or some kind of normal matrices), one can diagonalize them by unitary conjugations, which can be absorbed by -s in the exponent. Haar measure is invariant under such conjugations, so at the end these unitary conjugations are absorbed by the shift of the integration variable and we get exactly the same integral, but with and diagonal.
Fun fact: since and the (central) factor acts by
— it does nothing to , then integration over this part produces only some irrelevant constant factor and all the relevant part remains in the integration. I wouldn't distinguish between these domains of integration in what follows.
It possess character expansion [IZ80]
Character expansion derivation
Let's denote and expand exponent in Taylor series
so becomes
This integral looks quite complicated now. Well known integrals over unitary group are Schur orthogonality relations
so we will do our best to massage our integral into this form. One can notice that is nothing that power sum of matrix argument
where Young diagram is written in multiplicative notation. Power sums possess expansion in terms of Schur functions via inverse Frobenius characteristic map
where are characters of the representation of the symmetric group indexed by the partition evaluated at elements of cycle type indexed by the partition . The same map helps us to express through Schur function
Now we have
Great news! Schur functions are characters of the irreducible polynomial representation of (and hence, by restriction, of its compact form ). It means that
Now it's time for Schur orthogonality relations mentioned above
That's it!