Sokhotski–Plemelj theorem

Remarks

Proof idea: write $\frac{1}{x \pm i\varepsilon} = \frac{x}{x^2 + \varepsilon^2} \mp i \frac{\varepsilon}{x^2 + \varepsilon^2}$. The second kernel, divided by $\pi$, is an approximate identity, so its integral against $f$ tends to $\pi f(0)$. The first kernel, integrated against $f$, tends to the principal value; the Hölder condition makes $(f(x) - f(-x))/x$ integrable near $0$. In the language of distributions, $\frac{1}{x \pm i0} = \mathrm{p.v.} \frac{1}{x} \mp i\pi \delta$, where $\delta$ is the Dirac delta. Example: take $f = 1$ on $[-1, 1]$ and let $\operatorname{Log}$ be the principal logarithm. The first integral is $\operatorname{Log}(1 + i\varepsilon) - \operatorname{Log}(-1 + i\varepsilon)$, which tends to $0 - i\pi$, and the principal value is $0$. There is a version for Cauchy-type integrals over a curve, the Sokhotski–Plemelj formulae, which give the jump of $\frac{1}{2\pi i} \int_C \frac{\varphi(\zeta)}{\zeta - z} \, d\zeta$ across $C$.

Going deeper

For a student

The theorem gives the boundary values of Cauchy integrals, which is the starting point of the theory of singular integral equations and of the Riemann–Hilbert problem. In physics it appears whenever one adds $i\varepsilon$ to a denominator, as in Green's functions of quantum mechanics and in the Kramers–Kronig relations. It is also the computation that relates the Hilbert transform to boundary values of holomorphic functions on a half-plane.