= Solution
Write $D_n=\det[K(x_i,x_j)]_{i,j=1}^n$ and expand it by <permutations>. The assumptions ensure that the indicated single-variable integrals exist; the <finite-rank projection kernel> constructed above satisfies them with $r=m$. No symmetry of a general kernel is needed for the following argument.
If a <permutation> fixes $n$, its product contains the separate factor $K(x_n,x_n)$. Integration gives $r$ times the term of the corresponding <permutation> of $\{1,\ldots,n-1\}$. These terms contribute $rD_{n-1}$.
Otherwise $n$ belongs to a longer <permutation cycle>. Let $i$ precede $n$ and let $j$ follow it. The only factors containing $x_n$ are $K(x_i,x_n)K(x_n,x_j)$, whose integral is $K(x_i,x_j)$ by the reproducing assumption. Delete $n$ from the cycle to obtain a <permutation> $\tau$ on $n-1$ letters. Inserting $n$ back after any of the $n-1$ possible letters reconstructs exactly one <permutation>; its <sign of a permutation> is $-\operatorname{sgn}(\tau)$, since increasing a cycle length by one reverses its sign. Thus the nonfixed terms contribute $-(n-1)D_{n-1}$.
Combining these two disjoint classes proves <projection kernel determinant integration>:
$$
\boxed{\int_{\mathbb R}D_n\,dx_n=(r-n+1)D_{n-1}.}
$$
For $n=1$, use the empty <determinant> $D_0=1$, and the statement is precisely the assumed diagonal integral. For our kernel, <orthonormality> directly yields
$$
\int_{\mathbb R}K_m(x,y)K_m(y,z)\,dy=K_m(x,z),\qquad \int_{\mathbb R}K_m(x,x)\,dx=m,
$$
so every step of the marginal integration is justified.
Back to article page
