Categories: Complex analysis, Mathematics, Quantum mechanics.

Sokhotski-Plemelj theorem

The goal is to evaluate integrals of the following form, where f(x)f(x) is assumed to be continuous in the integration interval [a,b][a, b]:

lim⁡η→0+∫abf(x)x+iηdx\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} \end{aligned}

To do so, we start by splitting the integrand into its real and imaginary parts (limit hidden):

∫abf(x)x+iηdx=∫abx−iηx2+η2f(x)dx=∫abxx2+η2f(x)dx−i∫abηx2+η2f(x)dx\begin{aligned} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} &= \int_a^b \frac{x - i \eta}{x^2 + \eta^2} f(x) \dd{x} \\ &= \int_a^b \frac{x}{x^2 + \eta^2} f(x) \dd{x} - i \int_a^b \frac{\eta}{x^2 + \eta^2} f(x) \dd{x} \end{aligned}

In the real part, notice that the integrand diverges for x→0x \to 0 when η→0+\eta \to 0^+; more specifically, there is a singularity at zero. We therefore split the integral as follows:

lim⁡η→0+∫abxf(x)x2+η2dx=lim⁡η→0+(∫a−ηxf(x)x2+η2dx+∫ηbxf(x)x2+η2dx)\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} &= \lim_{\eta \to 0^+} \bigg( \int_a^{-\eta} \frac{x f(x)}{x^2 + \eta^2} \dd{x} + \int_\eta^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} \bigg) \end{aligned}

This is simply the definition of the Cauchy principal value P\mathcal{P}, so the real part is given by:

lim⁡η→0+∫abxf(x)x2+η2dx=P∫abxf(x)x2dx=P∫abf(x)xdx\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} &= \mathcal{P} \int_a^b \frac{x f(x)}{x^2} \dd{x} = \mathcal{P} \int_a^b \frac{f(x)}{x} \dd{x} \end{aligned}

Meanwhile, in the imaginary part, we substitute η\eta for 1/m1 / m, and introduce π\pi:

lim⁡η→0+∫abη f(x)x2+η2dx=lim⁡m→+∞ππ∫ab1/mx2+1/m2f(x)dx=lim⁡m→+∞ππ∫abm1+m2x2f(x)dx\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x} &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{1/m}{x^2 + 1/m^2} f(x) \dd{x} \\ &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{m}{1 + m^2 x^2} f(x) \dd{x} \end{aligned}

The expression m/π(1+m2x2)m / \pi (1 + m^2 x^2) is a so-called nascent delta function, meaning that in the limit m→+∞m \to +\infty it converges to the Dirac delta function δ(x)\delta(x):

lim⁡η→0+∫abη f(x)x2+η2dx=π∫abδ(x) f(x)dx=πf(0)\begin{aligned} \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x} &= \pi \int_a^b \delta(x) \: f(x) \dd{x} = \pi f(0) \end{aligned}

By combining the real and imaginary parts, we thus arrive at the (real version of the) Sokhotski-Plemelj theorem of complex analysis, which is important in quantum mechanics:

lim⁡η→0+∫abf(x)x+iηdx=P∫abf(x)xdx−iπf(0)\begin{aligned} \boxed{ \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} = \mathcal{P} \int_a^b \frac{f(x)}{x} \dd{x} - i \pi f(0) } \end{aligned}

However, this theorem is often written in the following sloppy way, where η\eta is defined up front to be small, the integral is hidden, and f(x)f(x) is set to 11. This awkwardly leaves P\mathcal{P} behind:

1x+iη=P1x−iπδ(x)\begin{aligned} \frac{1}{x + i \eta} = \mathcal{P} \frac{1}{x} - i \pi \delta(x) \end{aligned}

That was the real version of the theorem, which is a special case of a general result in complex analysis. Consider the following function:

ϕ(z)=∮Cf(ζ)ζ−zdζ\begin{aligned} \phi(z) = \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \end{aligned}

Where f(z)f(z) must be holomorphic. For all zz not on CC, this ϕ(z)\phi(z) exists, but not for z∈Cz \in C, since the integral diverges then. However, in the limit when approaching CC, we can still obtain a value for ϕ\phi, with a caveat: the value depends on the direction we approach CC from! The full Sokhotski-Plemelj theorem then states, for all zz on the closed contour CC:

lim⁡y→zϕ(y)=P∮Cf(ζ)ζ−zdζ± iπf(z)\begin{aligned} \boxed{ \lim_{y \to z} \phi(y) = \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \: i \pi f(z) } \end{aligned}

Where ±\pm is ++ if CC is approached from the inside, and −- if from outside. The above real version follows by making CC an infinitely large semicircle with its flat side on the real line: the integrand vanishes away from the real axis, because 1/(ζ ⁣− ⁣z)→01 / (\zeta \!-\! z) \to 0 for ∣ζ∣→∞|\zeta| \to \infty.