## Problem

The Sokhatsky-Weierstrass identity provides a simple formula for $1/(x+i\varepsilon)$, i.e. a simple pole that is shifted slightly to the lower half of the complex plane. It holds in the sense of distributions: $\lim\limits_{\varepsilon\rightarrow 0+}\frac{1}{x+i\varepsilon} = \mathcal{P}\frac{1}{x}-i\pi\delta(x)\,,$

where $\mathcal{P}$ stands for the Cauchy principal value. Proove it.

## Solution

In order to prove the identity, let us decompose $1/(x+i\varepsilon)$ into its real and imaginary part for finite $\varepsilon >0$, such that $\frac{1}{x+i\varepsilon} = \frac{x}{x^2+\varepsilon^2}-i\frac{\varepsilon}{x^2+\varepsilon^2}\,.$

We now consider both parts separately, starting with the real part. As stated before, the validity of the Sokhatsky-Weierstrass formula is understood as a distribution, i.e. we take an arbitrary test-function $f$, convolute it with the corresponding expression and apply the limit of vanishing $\varepsilon$ afterwards: $\lim\limits_{\varepsilon\rightarrow 0+}\int\limits_{-\infty}^\infty\frac{x}{x^2+\varepsilon^2}f(x)\,dx\,,$

where we set the upper and lower limits of the integration to $\pm\infty$ for simplicity. We split the range of integration into two regions, $|x|>\varepsilon$ and $|x|\leq\varepsilon$, such that $\int\limits_{-\infty}^\infty\frac{x}{x^2+\varepsilon^2}f(x)\,dx = \int\limits_{|x|\leq\varepsilon}\frac{x}{x^2+\varepsilon^2}f(x)\,dx+\int\limits_{|x|>\varepsilon}\frac{x}{x^2+\varepsilon^2}f(x)\,dx\,.$

The first term on the rhs. of this equation vanishes in the limit $\varepsilon\rightarrow 0$, since one can write $\int\limits_{|x|\leq\varepsilon}\frac{x}{x^2+\varepsilon^2}f(x)\,dx = \frac{1}{2}\int\limits_{-\varepsilon}^\varepsilon \frac{x}{x^2+\varepsilon^2}\Big(f(x)-f(-x)\Big)\,dx\,.$

Due to the mean value theorem there exist some number $\bar x$ inside $[-\varepsilon,\varepsilon]$ such that $\int\limits_{|x|\leq\varepsilon}\frac{x}{x^2+\varepsilon^2}f(x)\,dx = \varepsilon\frac{\bar x}{\bar x^2+\varepsilon^2}\Big(f(\bar x)-f(-\bar x)\Big)\,.$

Here the prefactor is finite in any case, since $\left| \frac{\varepsilon \bar x}{\bar x^2+\varepsilon^2}\right|\leq \frac{\varepsilon^2 }{\bar x^2+\varepsilon^2}\leq 1\,,$

on the other hand $\bar x\rightarrow 0$ as $\varepsilon\rightarrow 0$, thus the difference $\big(f(\bar x)-f(-\bar x)\big)$ vanishes in the same limit and so does the integral $\lim\limits_{\varepsilon\rightarrow 0+}\int\limits_{|x|\leq\varepsilon}\frac{x}{x^2+\varepsilon^2}f(x)\,dx = 0\,.$

In the second term on the rhs. of the real part one can expand the denominator in a geometric series, $\lim\limits_{\varepsilon\rightarrow 0+} \int\limits_{|x|>\varepsilon}\frac{x}{x^2+\varepsilon^2}f(x)\,dx = \mathcal{P}\int\limits_{-\infty}^\infty \frac{1}{x}f(x)\,dx + \lim\limits_{\varepsilon\rightarrow 0+}\sum\limits_{k=1}^\infty \int\limits_{|x|>\varepsilon} \frac{1}{x}\left(-\frac{\varepsilon^2}{x^2}\right)^kf(x)\,dx\,,$

where we already singled out the lowest contribution that is by very definition the principal value. One may (in principle) correctly conclude, that the other terms are of higher order in $\varepsilon$ and thus have to vanish when taking the limit $\varepsilon\rightarrow0$. To see this more explicitly let us rescale the integration variable $x\rightarrow x/\varepsilon$ to obtain $\lim\limits_{\varepsilon\rightarrow 0+}\sum\limits_{k=1}^\infty \int\limits_{|x|>\varepsilon} \frac{1}{x}\left(-\frac{\varepsilon^2}{x^2}\right)^kf(x)\,dx = \lim\limits_{\varepsilon\rightarrow 0+}\sum\limits_{k=1}^\infty \int\limits_{|x|>1} \frac{(-1)^k}{x^{2k+1}} f(\varepsilon x)\,dx\,,$

which vanishes by symmetry when replacing $f(\varepsilon x)\rightarrow f(0)$.

A similar trick can be done to obtain the imaginary part, using a suitable test-function $f$, $\lim\limits_{\varepsilon\rightarrow 0+}\int\limits_{-\infty}^\infty\frac{\varepsilon}{x^2+\varepsilon^2} f(x)\,dx = \lim\limits_{\varepsilon\rightarrow 0+}\int\limits_{-\infty}^\infty\frac{1}{\varepsilon}\frac{1}{(x/\varepsilon)^2+1} f(x)\,dx =\lim\limits_{\varepsilon\rightarrow 0+}\int\limits_{-\infty}^\infty\frac{1}{x^2+1} f(\varepsilon x)\,dx\,.$

Replacing $f(\varepsilon x)\rightarrow f(0)$ and the value of $\int\limits_{-\infty}^\infty\frac{dx}{x^2+1} = \pi$

yields the desired result.

From this one can also obtain the case of shifting the pole in the upper half of the complex plane and summarize both results as $\lim\limits_{\varepsilon\rightarrow 0+}\frac{1}{x\pm i\varepsilon} = \mathcal{P}\frac{1}{x}\mp i\pi\delta(x)\,.$

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