An identity of complete homogeneous symmetric polynomials $\sum_{k=1}^\infty k h_k(X)h_k(Y) t^{k-1}$

Question

Let $p,q$ be integers, $X, Y$ two groups of different variables, $X=(x_1,\ldots,x_p)$ and $Y=(y_1,\ldots,y_q)$, and $h_r$ a complete homogeneous symmetric polynomials. Is there an identity analogue to Cauchy's identity, in the form: $$ \sum_{k=1}^\infty k h_k(X)h_k(Y) t^{k-1}? $$

Answer 1

We know that $h_k(X)=\frac1{2\pi I}\oint \frac{dz}{z^{k+1}}\prod_i\frac{1}{1-z x_i}$, similarly for $h_k(Y)$. Then the sum is

$$ \frac{1}{(2\pi i)^2} \oint \oint \frac{wz\,dw dz}{(t-wz)^2} \prod_{i}\frac{1}{1-z x_i} \prod_{j}\frac{1}{1-w y_j}. $$

The integration contours are around $0$ and are such that $|wz|>t$.

I am not sure how to further simplify this expression.

Answer 2

We have $$ \sum_{k\geq 0} h_k(X)h_k(Y)t^k = \frac{1}{2\pi i} \int \frac{1}{\prod_i(1-sx_it)\prod_j(1-y_j/s)}\frac{ds}{s}, $$ integrated around $|s|=1$, where $x_i$ and $y_j$ are small. It can be shown by the Residue Theorem that this is equal (modulo computational error) to $$ \sum_j \frac{y_j^{q-1}}{\prod_i(1-x_iy_jt)\cdot \prod_{k\neq j}(y_j-y_k)}. $$ Now differentiate with respect to $t$. Not a very attractive answer, but I doubt that one can do better. Note that it is not clear from this formula that it is invariant under interchanging the $X$-variables with the $Y$-variables.