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Please tell me how to get the identity(2) in this image The derivation

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closed as off-topic by Jon Custer, Kyle Kanos, ZeroTheHero, stafusa, Buzz Dec 8 '18 at 2:27

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Draw a plot of the real and imaginary parts of

$$ \frac {1}{x-x_0-i\epsilon}=\frac{x-x_0}{(x-x_0)^2+\epsilon^2}+\frac{i\epsilon}{(x-x_0)^2+\epsilon^2} $$ Observe how the imaginary part sharpens to a delta function as $\epsilon\to 0$. Similarly observe how the real part provides a symmetrically cut-off version of $1/(x-x_0)$, which is the definition of $P(1/(x-x_0))$

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"Plot it" is unsatifactory in my view. I'm also not sure how you can deduce the real part is the principle value from looking at a plot. Here's a derivation. As we know, the delta function is only comfortably defined as a distribution - ie by the fact that \begin{align} \int f(x) \delta (x-x_0) dx = f(x_0) \end{align} So it's best that you think of the identity we're trying to prove as defined by what you get when you convolve the expression with a function. So consider the integral \begin{align} \int_{-a}^b \frac{f(x)}{x-i\epsilon}dx \end{align} The case of $+\epsilon$, or shifting $x\rightarrow x-x_0$ is a trivial generalisation. There's a pole at $x=i\epsilon$, above the real axis. Let's choose the contour, with a view that $\epsilon\rightarrow 0$:

enter image description here

The integral consists of three pieces. There is a piece along the negative real axis and one along the positive real axis: giving us the integral along the real axis, except that the singularity is avoided symmetrically around the origin. This contribution is given by \begin{align} \left(\int_{-a}^{-\epsilon}+\int_{\epsilon}^{b} \right)\frac{f(x)}{x} \ dx \end{align} This is by definition the principle value $\mathcal{P}\int_{-a}^b \frac{f(x)}{x} $. The remaining part of the integral is the arc below the pole. By taking the radius of the circle infinitesimally small, we can take $f(x)$ to be constant $f(0)$, and the integral along the semi circle is a half of that from a closed circle. By Cauchy's theorem, we get \begin{align} i\pi f(0)=\int i\pi f(x)\delta(x) dx \end{align} So the whole integral can be written \begin{align} \lim_{\epsilon\to 0}\int_{-a}^b \frac{f(x)}{x-i\epsilon} dx= \int_{-a}^b \mathcal{P}\ \frac{f(x)}{x} + i\pi\delta(x) f(x) \ dx \end{align} Taking real and imaginary parts gives your theorem. Performing the same integration around $x_0$ instead of the origin, or in the reverse direction for $+i\epsilon$, give the other usual generalisations.

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  • $\begingroup$ The contour integral is the probably best way if you know the technology. It does, though, require that $f(x)$ be analytic. The P+ $i$ delta formula holds even if $f(x)$ is not analytic, and recognizing what the plot is doing provides the key a rigorous proof of the fact that $$\lim_{\epsilon\to 0}\frac{x}{x^2+\epsilon^2}= P(1/x)$$ as a distribution.. $\endgroup$ – mike stone Dec 13 '18 at 22:57

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