The trace of an integral equation?

Question

I am reading a paper about spectroanalysis and encountered the following integral equation: $$\int_{-1}^{1}\frac{\sin A(x-x')}{\pi(x-x')}\psi(x')dx'=\lambda\psi(x)$$ Then the paper gives without proof $$\sum_{\alpha}\lambda_{\alpha}={2A\over \pi}$$ Can it be viewed as the trace of this integral equation? And how can it be derived?

Answer 1

The idea is that an integral operator $Tf(x)=\int_a^b K(x,x')\,f(x')\;dx'$ with $K(x,x')=\overline{K(x',x)}$ has trace $\int_a^b K(x,x)\;dx$, that is, integrating down the diagonal, as though one were taking trace of a matrix.

Operators given by genuinely symmetric, smooth kernels on compact intervals in one dimension are provably trace-class, so this is legit.

That is, the trace evaluated in this way is equal to the sum of eigenvalues.

In the case at hand, the diagonal values are the constant $A/\pi$, and the length of the interval is $2$, so one has the assertion.