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I have read the answer to this question and it has helped a lot but for$\int_{-\infty}^{\infty}\frac{1}{x-2}dx$, I get the wrong answer. I used this contour \begin{align} P.V\int_{-\infty}^{\infty}\frac{1}{x-2}dx&=\int_{C\ \cup P\ \cup L_1\cup L_2}^{}\frac{1}{z-2}dz-\int_{C}^{}\frac{1}{z-2}dz-\int_{P}^{}\frac{1}{z-2}dz\\ &=0-0-(-i\pi)=i\pi \end{align} As you can see the answer is wrong because it's complex even though it's the result of a real integral. And I suppose that we can't be using $P.V=\text{Re}(i\pi)$ because there's no $\cos(ax)$ in the numerator.

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  • $\begingroup$ You seem to be assuming that $\int_C\to0$ as the radius tends to $\infty$. Just because it works that way in the examples you've seen doesn;t mean it works here - how did you show $\int_C\to0$??? $\endgroup$ Jan 9 at 15:02
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    $\begingroup$ Btw using complex to find this PV seems silly. What is $\int_{2-R}^{2-\delta}+\int_{2+\delta}^{2+R}$? $\endgroup$ Jan 9 at 15:04
  • $\begingroup$ @DavidC.Ullrich Yes I guess you're right, I probably have to use some other lemma to determine the value of the integral, and yes it's much easier to find the PV but I just wanted to learn the method with different examples. $\endgroup$
    – YetiAsk
    Jan 9 at 17:13
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Your problem comes from the fact that the integral over $C$ is wrong. It not zero.

Suppose that $C$ were a semicircle given by $z-2=100\exp(i\theta), 0\le\theta\le\pi$. You evaluate the integral over this path, what do you get? Can you see that the result will be independent of the radius of the large semicircle?

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