# How to deal with poles along contour of integration

Suppose one has an integral of a real (or complex) function that has at least one pole that lies on the contour of integration, for example: $$\int_{-a}^{a}\frac{f(x)}{x}dx$$ Clearly such an integral is undefined in the normal sense.

My question is, how does one deal with such integrals in general? I've heard of the notion of a Cauchy Principal value, but not sure I fully understand the method. Does one deform the contour slightly to avoid the pole(s) and then take the limit as this deformation tends to zero, i.e. $$\mathcal{P}\int_{-a}^{a}\frac{f(x)}{x}dx=\lim_{\delta\rightarrow 0}\left(\int_{-a}^{-\delta}\frac{f(x)}{x}dx+\int_{\delta}^{a}\frac{f(x)}{x}dx\right)\;\text{?}$$

I'm approaching this from a physicist's perspective, in particular in relation to the Feynman propagator, which as an integral along the real axis is not well-defined since the integrand has two poles lying on the real axis. In the standard approach the contour is deformed such that it does not pass through the poles, and then the residue theorem is applied. However, the issue of whether this converges to the actual integral once the limit is taken is usually glossed over. In a more technical approach, is one technically taking the Cauchy principal value of the original integral?

• In your example: $P \int_{-a}^a f(x)/x\equiv\lim_{\epsilon\rightarrow0}\int_{-a}^{\epsilon} f(x)/x+\int_{\epsilon}^a f(x)/x$. This definition induces some nice properties for example, if $f(x)$ even the Principal value vanishs – tired May 21 '17 at 18:05
• From a complex analyis perspective, your considerations are more or less correct : $$P \int_{-a}^a f(x)/x=-\pi i f(0)-\int_{closure}f(z)dz$$. Here i assumed that $f(x)$ is analytic in the upper half plane and $closure$ is some path that closes the contour of integration so that we can apply residue theorem – tired May 21 '17 at 18:09
• @tired In terms of the Feynman propagator, is one technically taking the Cauchy principal value of the associated contour integral? – user35305 May 21 '17 at 18:20
• @op more or less, you have to include additionally the contributions from the small indents around the poles of your propagator in k-space. If my brain serves well, there are two of them in the case of Feynman's Propagator for a free (or nicely renormalized) quantum field which might cancel due to symmetries of the underlying physical system – tired May 21 '17 at 18:25
• In a Feynman propagator, you never take the Cauchy principle value blindly. Instead, you move the contour around the pole. The small shift of the contour in the imaginary direction corresponds to regularizing the integral in a certain manner. Remember, everything in physics is approximation. The regularization you employ should reflect the difference between the real world and the ideal equation you write down to model the system. This usually dictates which direction you should move the contour. – achille hui May 21 '17 at 18:39