# Principal value of $\int^\infty_{-\infty}\frac1{x^2}dx$: counter-intuitive?

In attempt to evaluate $$\text P\int^\infty_{-\infty}\frac1{x^2}dx$$ we consider $$\oint_C\frac{1}{z^2}dz$$ where $C$ is an infinitely large semicircle on the upper half plane centered at the origin, with a small indent at the origin.

Obviously, the indent integral and the large arc integral tends to zero. By Cauchy’s integral theorem, the contour integral is also zero. Thus, $$\text P\int^\infty_{-\infty}\frac1{x^2}dx=0$$

But the integrand is always positive. How could the integral equal zero? How can this counter-intuitive result be explained?

• $1/z^2$ is not continuous on the semi disk. – xbh Aug 19 '18 at 6:29
• The indent integral does not tend to zero. – pks Aug 19 '18 at 6:52
• I don't even see any relations between the Cauchy integral theorem and the principal value, yet. Where does your conclusion of the principal value come from? Anyway, the principal value of $\frac{1}{x^2}$ does not exist using its definition since you already determined that the integrand is always positive. – YoungMath Aug 19 '18 at 9:49

The issue is first in your obvious step, which then makes it wrong to take the limit after applying Cauchy's theorem. The (counterclockwise) integral over the small semicircle is $$\int_{\partial B(0,\epsilon)\cap \{\Im z>0\}} \frac {dz}{z^2} = \int_0^\pi \frac{i \epsilon e^{i\theta}}{\epsilon^2 e^{i2\theta}} \ d\theta = \frac 1\epsilon \int_0^\pi ie^{-i\theta} d\theta = \frac{2}\epsilon \to \infty$$
However you might note that the (counterclockwise) integral over the larger semicircle is similarly $\int_{\partial B(0,R)\cap \{\Im z>0\}} = \frac2R$, and on the line segments we have $$\int_{[-R,R]\setminus[-\epsilon,\epsilon]} \frac{dz}{z^2} = 2\int_{\epsilon}^R \frac{dx}{x^2} = \frac2{\epsilon}- \frac2{R}$$