Integrating a Function with a Removable Singularity

Back to Contents

In this example we want to examine the integral of (Sinx)/x evaluated over the range from zero to infinity. That function appears to have a singularity at x=0, but because both Sinx and x go to zero together it=s actually an indeterminate point, one that we clear up by laying out the function=s Taylor series expansion;

(Eq=n 1)

That expansion tells us that at the point x=0 the function equals one, so it doesn=t represent a true singularity at all. We can integrate that equation term by term to get

(Eq=n 2)

So far no one has found a closed-form solution equal to that infinite series, so that integral will be extremely tedious to calculate. However, we can easily evaluate one definite version of that integral.

We want to evaluate

(Eq=n 3)

Since we want to integrate our given function entirely on the real axis, we can complexify the integrand without changing the value of the integral. So we have now

(Eq=n 4)

in which Im indicates that we want to keep only the imaginary part of the integral, the sine part. But that integral has a real singularity, a simple pole at z=0 due to the cosine component of the imaginary exponential, which means that we can now use Cauchy=s integral theorem directly in solving the integral. Because our original integrand is symmetrical about the point x=0 and because we don=t want to take our integration off the real-number line until we have gone infinitely far from the origin of our complex plane, we take our integral between the limits plus infinity and minus infinity;

(Eq=n 5)

Now we want to use Cauchy=s integral theorem to integrate our function over a half disc lying entirely in the upper half of the complex plane (all y>0), centered on the origin, and excluding a minuscule half disc at the origin. We take R to represent the larger radius of the half disc and r to represent the smaller radius of the half disc (the radius of the excluded half disc). Thus we have

(Eq=n 6)

In that equation we have

(Eq=n 7)

in which we describe the semicircle (πr) on which we integrate as z=reiθ and note that in the limit as r goes to zero the exponential goes to one.

That neat little trick won=t work for the integral on the larger semicircle, the one whose radius goes to infinity. For that integral we must carry out the actual integration. If we integrate repeatedly by parts, we will get an infinite series that will do what we need. We have

(Eq=n 8)

As we did with the small semicircle, we describe the large semicircle by setting z=Reiθ. When we substitute that into Equation 8 we can see that the exponentials won=t exceed one and won=t go to zero. Thus, when we let R go toward infinity, the value of the integral goes toward zero, as the limit requires.

Rewriting Equation 6 gives us

(Eq=n 9)

Referring back to Equations 3, 4, and 5, we can see that equation giving us

(Eq=n 10)

which we originally wanted, and, incidentally,

(Eq=n 11)

Example Application:

Ferroelectric materials are the electric analogues of ferromagnetic materials: when immersed in an electric field they acquire a permanent electric polarization just as ferromagnetic materials immersed in a magnetic field acquire a permanent magnetization. Over very short intervals of time the rate at which a ferroelectric material polarizes stands in direct proportion to the strength of the applied electric field;

(Eq=n 12)

Imagine that a small block of ferroelectric material floats in space and that we illuminate it with a radio wave in which the amplitude of the electric field declines uniformly with the elapse of time (E=E0/βt). The strength of the electric field acting on the block at any instant then conforms to

(Eq=n 13)

Over any minuscule interval dt the block=s polarization changes by

(Eq=n 14)

After a sufficiently long interval has elapsed, the radio wave has diminished effectively to zero and the net polarization that we expect to find in the block conforms to

(Eq=n 15)

efefabefef

Back to Contents