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diff --git a/content/know/concept/sokhotski-plemelj-theorem/index.pdc b/content/know/concept/sokhotski-plemelj-theorem/index.pdc new file mode 100644 index 0000000..a482618 --- /dev/null +++ b/content/know/concept/sokhotski-plemelj-theorem/index.pdc @@ -0,0 +1,115 @@ +--- +title: "Sokhotski-Plemelj theorem" +firstLetter: "S" +publishDate: 2021-11-01 +categories: +- Mathematics +- Complex analysis +- Quantum mechanics + +date: 2021-11-01T12:54:37+01:00 +draft: false +markup: pandoc +--- + +# Sokhotski-Plemelj theorem + +The goal is to evaluate integrals of the following form, where $a < 0 < b$, +and $f(x)$ is assumed to be continuous in the integration interval $[a, b]$: + +$$\begin{aligned} + \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} +\end{aligned}$$ + +To do so, we start by splitting the integrand +into its real and imaginary parts (limit hidden): + +$$\begin{aligned} + \int_a^b \frac{f(x)}{x + i \eta} \dd{x} + &= \int_a^b \frac{x - i \eta}{x^2 + \eta^2} f(x) \dd{x} + = \int_a^b \bigg( \frac{x}{x^2 + \eta^2} - i \frac{\eta}{x^2 + \eta^2} \bigg) f(x) \dd{x} +\end{aligned}$$ + +To evaluate the real part, +we notice that for $\eta \to 0^+$ the integrand diverges for $x \to 0$, +and thus split the integral as follows: + +$$\begin{aligned} + \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} + &= \lim_{\eta \to 0^+} \bigg( \int_a^{-\eta} \frac{x f(x)}{x^2 + \eta^2} \dd{x} + \int_\eta^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} \bigg) +\end{aligned}$$ + +This is simply the definition of the +[Cauchy principal value](/know/concept/cauchy-principal-value/) $\mathcal{P}$, +so the real part is given by: + +$$\begin{aligned} + \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} + &= \mathcal{P} \int_a^b \frac{x f(x)}{x^2} \dd{x} + = \mathcal{P} \int_a^b \frac{f(x)}{x} \dd{x} +\end{aligned}$$ + +Meanwhile, in the imaginary part, +we substitute $\eta$ for $1 / m$, and introduce $\pi$: + +$$\begin{aligned} + \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x} + &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{1/m}{x^2 + 1/m^2} f(x) \dd{x} + \\ + &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{m}{1 + m^2 x^2} f(x) \dd{x} +\end{aligned}$$ + +The expression $m / \pi (1 + m^2 x^2)$ is a so-called *nascent delta function*, +meaning that in the limit $m \to +\infty$ it converges to +the [Dirac delta function](/know/concept/dirac-delta-function/): + +$$\begin{aligned} + \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x} + &= \pi \int_a^b \delta(x) \: f(x) \dd{x} + = \pi f(0) +\end{aligned}$$ + +By combining the real and imaginary parts, +we thus arrive at the (real version of the) +so-called **Sokhotski-Plemelj theorem** of complex analysis: + +$$\begin{aligned} + \boxed{ + \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x} + = \mathcal{P} \int_a^b \frac{f(x)}{x} \dd{x} - i \pi f(0) + } +\end{aligned}$$ + +However, this theorem is often written in the following sloppy way, +where $\eta$ is defined up front to be small, +the integral is hidden, and $f(x)$ is set to $1$. +This awkwardly leaves $\mathcal{P}$ behind: + +$$\begin{aligned} + \frac{1}{x + i \eta} + = \mathcal{P} \Big( \frac{1}{x} \Big) - i \pi \delta(x) +\end{aligned}$$ + +The full, complex version of the Sokhotski-Plemelj theorem +evaluates integrals of the following form +over a contour $C$ in the complex plane: + +$$\begin{aligned} + \phi(z) = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} +\end{aligned}$$ + +Where $f(z)$ must be [holomorphic](/know/concept/holomorphic-function/). +The Sokhotski-Plemelj theorem then states: + +$$\begin{aligned} + \boxed{ + \lim_{w \to z} \phi(w) + = \frac{1}{2 \pi i} \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \frac{f(z)}{2} + } +\end{aligned}$$ + +Where the sign is positive if $z$ is inside $C$, and negative if it is outside. +The real version follows by letting $C$ follow the whole real axis, +making $C$ an infinitely large semicircle, +so that the integrand vanishes away from the real axis, +because $1 / (\zeta \!-\! z) \to 0$ for $|\zeta| \to \infty$. |