wrong answer in calculating the principal value of integral using complex analysis

Question

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.

Answer 1

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?