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By definition, $$ \begin{align} \text{PV} \int_{-\infty}^{\infty} \frac{1}{x^{3}} \ dx &= \lim_{\epsilon \to 0^{+}} \left(\int_{- 1/ \epsilon}^{- \epsilon} \frac{1}{x^{3}} \ dx + \int_{ \epsilon}^{1 / \epsilon} \frac{1}{x^{3}} \ dx \right) \\ &=\lim_{\epsilon \to 0^{+}} \left(-\int_{\epsilon}^{1 / \epsilon} \frac{1}{x^{3}} \ dx + \int_{\epsilon}^{1 / \epsilon} \frac{1}{x^{3}} \ dx \right) \\ &=0. \end{align}$$

But how do you find the Cauchy principal value using contour integration?

If you let $ \displaystyle f(z) = \frac{1}{z^{3}}$ and integrate around an indented contour that consists of the real axis and the upper half of $|z|=R$, won't the pole of order 3 at the origin cause everything to blow up?

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Integrate over a large semi-circle with a tiny indentation around $0$. The integral of $f$ along every such contour is $0$, and the part coming from the large semi-circle will tend to $0$ as the radius tends to $+\infty$.

Finally the integral over the tiny indentation will be $0$ for every $\varepsilon > 0$, either by parametrizing or by using a primitive for $f$ (and this is despite $f$ having a triple pole at the origin.) But as you say, you won't be able to show this by just estimating $|f|$ on the indentation.

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    $\begingroup$ Wow. It evaluates to zero even before taking the limit. I wasn't expecting that. So for odd negative integer powers (i.e., $\frac{1}{z^{2n+1}}$) it will contribute nothing, but for even negative integer powers it will blow up there in the limit? $\endgroup$ – Random Variable Mar 6 '13 at 23:57
  • $\begingroup$ @RandomVariable yes. $\endgroup$ – mrf Mar 7 '13 at 0:06

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