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With a difficult integral is it possible to split it up and then apply the Cauchy principal value theorem to the first integral to turn it to $0$, then to use contour integration on the other.

It just seems dodgy to use CPV on one term then contour integration on the other, although the split integrals have different numbers of singularities.

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It would be easier to provide help if you gave a specific example of an integral that you care about. –  Greg Martin Mar 26 '13 at 16:09
    
in principle, yes (at least in physics problems). but as Greg said, it would be easier if you gave the specific integral. –  lomppi Mar 26 '13 at 16:58
    
The integral is $$\int^{+\infty}_{-\infty} x + \frac{1}{x} dx$$ so CPV on x dx gives 0, then contour integration on $\int 1/x dx$ gives $i\pi$ as an answer –  user63407 Mar 26 '13 at 18:22
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2 Answers 2

I think this calculation can be somewhat justified without splitting the function.

The integral $\int_{-\infty}^\infty (x+x^{-1})\,dx$ has problems at $0$ and at $\infty$. Your question concerns two ways of getting around problematic points, where a function is not integrable:

  1. CPV: remove a symmetric neighborhood of the problematic point from the region of integration. Then let its "radius" tend to $0$.
  2. Contour integration: replace a symmetric neighborhood of the problematic point with a semicircle. Then let its "radius" tend to $0$.

The concept of "radius" must be understood differently when the center is $\infty$: let's say that the circle of radius $r$ around $\infty $ is $\{|z|=1/r\}$.

$$\int_{|x|\le 1} (x+x^{-1})\,dx+ \int_{|x|\ge 1} ( x+x^{-1})\,dx$$

For the first integral, going around $0$ through the complex plane brings in $\pi i$ (or $-\pi$, depending on how you do it). For the second one, replacing a neighborhood of infinity with semi-circle does not work very well: the integral over semi-circle has no finite limit either. Perhaps this is why the CPV approach is used, which makes $\int_{|x|\ge 1} ( x+x^{-1})\,dx=0$.


To actually evaluate the appropriateness of such manipulations, one needs more context.

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Personally I don't quite see the problem? It's perfectly possible to use a different technique on the two integrals:once you've split it using the fact that it's a sum of two functions in x, then the two parts are independent and you can integrate or simplify either of them any way you like!

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