# Principal values (of integrals)

I’ve been looking through my Gamelin’s Complex Analysis quite a bit lately. I’ve solved some exercises, which I’ve written up in private. I was just going over the section on principal values, which had a very neat calculation. I’ll give a sketch of that one here.

Take an integral $\int_a^b f(x)dx$ such that on some $x_0 \in (a,b)$ there is a singularity, such as $\int_{-1}^1 \frac{1}{x}dx$. The principal value of that is defined as $PV \int_a^b f(x)dx = \lim_{\epsilon \to 0}\left(\int_a^{x_0 - \epsilon} + \int_{x_0 + \epsilon}^b\right)f(x)dx$.

The example the book presented was $PV\int_{-\infty}^{\infty} \frac{1}{x^3 - 1} = -\frac{\pi}{\sqrt{3}}$.

Its calculation invokes both the residue theorem and the fractional residue theorem. Our integrand, complexly viewed, has a singularity at $e^{2\pi i / 3}$, with residue $\frac{1}{3z^2}|_{z = e^{2\pi i / 3}} = \frac{e^{2\pi i / 3}}{3}$, which one can arrive at with so called Rule 4 in the book, or more from first principles, l’Hopital’s rule. That is the residue to calculate if we had the half-disk in the half plane, arbitrarily large. However, with our pole at $1$ we must indent it there. The integral along the arc obviously vanishes. The infinitesimal arc spawned by the indentation, the integral along which, can be calculated by the fractional residue theorem, with any $-\pi$, the minus accounting for the clockwise direction. This time the residue is at $1$, with $\frac{1}{3z^2}|_{z = 1} = \frac{1}{3}$. So that integral, no matter how small $\epsilon$ is, is $-\frac{\pi}{3}i$. $2\pi i$ times the first residue we calculated minus that, which is extra with respect to the integral, the principal value thereof, that we wish to calculate, yields $-\frac{\pi}{\sqrt{3}}$ for the desired answer.

Let’s generalize. Complex analysis provides the machinery to compute integrals not to be integrated easily by real means, or something like that. Canonical is having the value on an arc go to naught as the arc becomes arbitrarily large, and equating the integral with a constant times the sum of the residues inside. We’ve done that here. Well, it turns out that if the integral has an integrand that explodes somewhere on the domain of integration, we can make a dent there, and minus out the integral along its corresponding arc.