## Cauchy Principal Value

On Math.SE, we frequently get integrals that simply do not exist in the usual sense because the integration path intersects a pole of the integrand. Occasionally, we come across such integrals in the course of evaluating integrals of functions with removable singularities using complex methods. However, sometimes such integrals are interesting in their own right. For example, the complex permittivity of a medium obeys something called the Kramers-Kronig relations in which the real part of the permittivity as a function of frequency is expressed as an integral of the imaginary part over all frequencies. However, the integrand of the integral has a pole at a frequency of interest. (See J.D. Jackson, Classical Electrodynamics, 3rd Ed., Chap. 7.11). Thus, there is a desire to define such integrals in such a way that the singular part due to the pole may be removed.

The Cauchy Principal value of an integral of a function $f(x)$ that has a single pole at $x=a$ is defined to be

$$PV \int_{-\infty}^{\infty} dx \, f(x) = \lim_{\epsilon \to 0} \left [\int_{-\infty}^{a-\epsilon} dx \, f(x) + \int_{a+\epsilon}^{\infty} dx \, f(x)\right ]$$

Of course, in many cases the limit may be evaluated directly. However, there are many cases in which the limit may be evaluated indirectly using complex integration. (Surprise surprise.)

Recently, I worked an example that beautifully illustrated the benefits of such an evaluation in the complex plane. The problem is as follows: evaluate the integral

$$I = \int_{-\infty}^{\infty} dx \frac{e^{-\frac{x^{2}}{2}}}{x – a – i 0}$$

The notation $i 0$ is in my opinion a rather unfortunate way to express a Cauchy principal value integral, but what the problem essentially states is that the pole is to be considered as having an infinitesimally small imaginary part in the upper half plane. This is typically an important detail because quantities like the complex permittivity are analytic in the upper half plane (a consequence of causality in the time domain).

$$PV \int_{-\infty}^{\infty} dx \frac{e^{-x^2/2}}{x-a} = 2 a \, PV \int_{0}^{\infty} dx \frac{e^{-x^2/2}}{x^2-a^2}$$

Note that the latter integral is over all positive numbers. This allows us a much more convenient representation of the integral in the complex plane.

Along these lines, consider the contour integral

$$\oint_C dz \frac{e^{-z^2/2}}{z^2-a^2}$$

where $C$ is a wedge of angle $\pi/4$ of radius $R$ in the upper right quadrant, with a semicircular detour of radius $\epsilon$ into the upper half plane at $z=a$. The contour integral is equal to

$$PV \int_{0}^{R} dx \frac{e^{-x^2/2}}{x^2-a^2} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{-(a+\epsilon e^{i \phi})^2/2}}{(a+\epsilon e^{i \phi})^2-a^2} \\ + i R \int_0^{\pi/4} d\theta \, e^{i \theta} \frac{e^{-R^2 e^{i 2 \theta}/2}}{R^2 e^{i 2 \theta}-a^2}+ e^{i \pi/4} \int_R^0 dt \, \frac{e^{-i t^2/2}}{i t^2-a^2}$$

Take the limit as $R \to \infty$ and $\epsilon \to 0$. The third integral vanishes in this limit. The second integral approaches $-i \pi/(2 a) e^{-a^2/2}$.

By Cauchy’s theorem, the contour integral is zero. Thus, we can say that

$$PV \int_{-\infty}^{\infty} dx \frac{e^{-x^2/2}}{x-a} = i \pi \, e^{-a^2/2} – 2 a e^{i \pi/4} \int_0^{\infty} dt \, \frac{e^{-i t^2/2}}{i t^2-a^2}$$

The integral is almost straightforward to evaluate. Write the integral as

$$e^{-y a^2}\int_0^{\infty} dt \, \frac{e^{-y (i t^2-a^2)}}{i t^2-a^2}$$

where $y=1/2$. Now define

$$I(y) = \int_0^{\infty} dt \, \frac{e^{-y (i t^2-a^2)}}{i t^2-a^2}$$

$$I'(y) = -e^{y a^2} \int_0^{\infty} dt \, e^{-i y t^2} = -\frac12 \sqrt{\pi} e^{-i \pi/4} y^{-1/2} e^{y a^2}$$

$$I(0) = \int_0^{\infty} \frac{dt}{i t^2-a^2} = -i e^{-i \pi/4} \frac{\pi}{2 a}$$

Then

$$I(y) = -i e^{-i \pi/4} \frac{\pi}{2 a} – \frac12 \sqrt{\pi} e^{-i \pi/4} \int_0^y dy’\, y’^{-1/2} e^{y’ a^2} = -i e^{-i \pi/4} \frac{\pi}{2 a} – \frac{\pi}{2 a} e^{-i \pi/4} \operatorname{erfi}{\left (a \sqrt{y}\right )}$$

where $\operatorname{erfi}$ is the imaginary error function

$$\operatorname{erfi}{x} = \frac{2}{\sqrt{\pi}} \int_0^x dt \, e^{t^2}$$

Finally, we may plug in $y=1/2$ and we have

$$PV \int_{-\infty}^{\infty} dx \frac{e^{-x^2/2}}{x-a} = -\pi e^{-a^2/2} \operatorname{erfi}{\left (\frac{a}{\sqrt{2}}\right )}$$

This result has a really nice physical interpretation. Let’s say that the real part of a complex permittivity has the form

$$\operatorname{Re}{[n^2(\omega)-1]} = e^{-\omega^2/2}$$

Then the imaginary part of the complex permittivity is, by the Kramers-Kronig relation:

$$\operatorname{Im}{[n^2(\omega)-1]} = \operatorname{Im}{[n^2(\omega)]}= e^{-\omega^2/2} \operatorname{erfi}{\left (\frac{\omega}{\sqrt{2}}\right )}$$

Here is a plot of the real and imaginary parts of the permittivity:

From this plot, the global dispersive and absorptive properties of this medium may be deduced. Really amazing stuff given simple causality requirements!

• Huang Phillip wrote:

Hi,

thanks very much for the illustration. I am now working on a similar integral and your illustration is very useful.

One question is: how do you evaluate the integral on the semicircular detour of radius ϵ? It involvs to integrate a term like: int(exp(exp(ix)))dx. Can you give some hint for how to deal with the integral of a exponential of a exponential component in the complex plane?

• Huang Phillip wrote:

maybe I should first separate the real and imaginary part of exp(ix)?

• Phillip: The term typically looks like $\exp{ \epsilon \exp{i \phi}}$; in that case, one would expand the outer exponential for small $\epsilon$.

• Huang Phillip wrote:

yes , it works. Thanks Ron

• Mahmoud Elzouka wrote:

Thanks for the elaborate illustration. You have saved me around an hour 🙂