I have an embarrassingly basic question about the Szegő kernel on the Hardy space $H^2$ over the right half plane. It seems I have forgotten as much complex analysis as I ever knew...
A kernel $k(s, z)$ is said to satisfy the reproducing property if $$ \langle f, k(s, \cdot) \rangle = f(s). $$
On $H^2$, the inner product is $$ \langle f, g \rangle = \int_{-\infty}^{\infty} f(j\omega) \overline{g(j\omega)}d\omega, $$
and $k(s, z)$ is known to be the Szegő kernel, $1/(2\pi(z + \bar s))$.
My question is: how does one prove that $k(s, z)$ has the reproducing property? Plugging into the inner product, I want to show that $$ \int_{-\infty}^{\infty} \frac{f(j\omega)}{2\pi(s - j\omega)}d\omega = f(s). $$ I guess this follows from the residue theorem, but I can't see why the residue of $f(z)/(2\pi(s - \bar z))$ is $f(s)/2j\pi$...
Edit: it seems the only issue was that I wasn't orienting the boundary of the right half plane positively (counter-clockwise). By the residue theorem: $$ 2\pi j \operatorname{Res}\left(\frac{f(z)}{z-s}\right) = \int_{\partial \mathbb{C}^+} \frac{f(j\omega)}{j\omega - s}d(j\omega)\\ f(s) = \frac{1}{2\pi}\int_\infty^{-\infty} \frac{f(j\omega)}{j\omega - s} d\omega\\ = \int_{-\infty}^{\infty} \frac{f(j\omega)}{2\pi(s - j\omega)} d\omega\\ = \langle f, k(s, \cdot) \rangle. $$