Determinantal formula for reproducing integral kernel How do I prove the following?
$$\int\det\left(K(x_{i},x_{j})\right)_{1\leq i,j\leq n}dx_{1} \cdots dx_{N}=\underset{i=1}{\overset{n}{\prod}}\left(\int K(x_{i},x_{i})\;dx_{i}-(i-1)\right)$$
where
$$K(x,y)=\sum_{l=1}^n \psi_l(x)\overline{\psi_l}(y)$$
and
$$\{\psi_l(x)\}_{l=1}^n$$
is an ON-sequence in $L^2$.
One may note that $$\int K(x_i,x_j)K(x_j,x_i) \; d\mu(x_i)=K(x_j,x_j)$$
and also that $$\int K(x_a,x_b)K(x_b,x_c)d\mu(x_b)=K(x_a,x_c).$$

(This is too long to fit into a comment.)

Note that the product in the integrand of the right side foils out as
$$\sum_{A\subseteq [n]} (-1)^{n-|A|}\left(\prod_{i\not\in A} (i-1)\right)\prod_{j\in A}K(x_j,x_j).$$
(Here $[n]=\{1,2,\dots,n\}$.) On the other hand, we can use Leibniz formula to expand the determinant in the left hand side to obtain
$$\iint\cdots\int \sum_{\sigma\in S_n} (-1)^{\sigma}\left(\prod_{i=1}^nK(x_i,x_{\sigma(i)}) \right)dx_1\cdots dx_n$$
Interchange the order of summation and integration. Observe that $K(x_1,x_{\sigma(1)})\cdots K(x_n,x_{\sigma(n)})$ can be reordered into a product of factors of the form $K(x_a,x_b)K(x_b,x_c)\cdots K(x_d,x_a)$ (this follows from the cycle decomposition of the permutation $\sigma$), and the integration of this over $x_b,x_c,\dots,x_d$ is, by induction on the two properties at the bottom of the OP, smply $K(x_a,x_a)$. From here we can once again switch the order of summation and integration.
I think the last ingredient needed is something from the representation theory of the symmetric group. In other words, we need to know that the number of ways a permutation $\sigma$ can be decomposed into cycles with representatives $i_1,i_2,\dots\in A\subseteq [n]$, weighed by sign, is the coefficient of $\prod_{j\in A}K(x_j,x_j)$ in the integrand's polynomial on the right hand side (or something roughly to this effect).
