Cauchy principal value

Question

The author of my old complex analysis textbook frequently asks the reader to calculate the Cauchy principal values of absolutely convergent real-valued integrals, e.g., EDIT: $\displaystyle\text{PV}\int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx $. For a long time I thought that meant that EDIT: $\displaystyle\text{PV}\int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx \ne \int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx $. But that's not correct, right?

Answer 1

Generally PV is used in a "two-sided" situation: $PV \int_{-\infty}^\infty$ (where it means the limit of $\int_{-R}^R$ as $R \to +\infty$), or $PV \int_a^b$ if there is a singularity at $c \in (a,b)$ (where you take the limit of $\int_a^{c-\epsilon} + \int_{c+\epsilon}^b$ as $\epsilon \to 0+$).

In elementary complex analysis you most often encounter $\int_{-\infty}^\infty$. $PV \int_0^\infty \frac{\cos x}{1+x^2}\ dx$ doesn't really make sense as a principal value integral, but it actually comes from $PV \int_{-\infty}^\infty \frac{\cos x}{1+x^2} \ dx$, where the PV does make sense, by using symmetry. Of course the PV is not necessary here because the integral converges absolutely, and that will occur in very many of the examples.