The author of my complex analysis textbook asks the reader to find the Cauchy principal value of absolutely convergent real-valued integrals such as $\displaystyle\int_{{\color{red}{-\infty}}}^\infty \frac{\cos x}{1+x^{2}} \ dx $.
For a long time I thought that meant that $$\text{PV}\int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx = \lim_{R \to \infty} \int_{-R}^{R} \frac{\cos x}{1+x^{2}} \ dx \ne \int_{-\infty}^\infty \frac{\cos x}{1+x^{2}} \ dx. $$
But doesn't an absolutely convergent real-valued integral equal its Cauchy principal value?
Best Answer
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.