# Dispersion Relations in Particle Physics [closed]

Please tell me how to get the identity(2) in this image ## closed as off-topic by Jon Custer, Kyle Kanos, ZeroTheHero, stafusa, BuzzDec 8 '18 at 2:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Jon Custer, Kyle Kanos, ZeroTheHero, stafusa
If this question can be reworded to fit the rules in the help center, please edit the question.

$$\frac {1}{x-x_0-i\epsilon}=\frac{x-x_0}{(x-x_0)^2+\epsilon^2}+\frac{i\epsilon}{(x-x_0)^2+\epsilon^2}$$ 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 $$P(1/(x-x_0))$$
"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 $$\epsilon\rightarrow 0$$: 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.
• 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.. – mike stone Dec 13 '18 at 22:57