Singularities and Their Residues

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If we want to use Cauchy=s integral theorem to evaluate an integral in the complex plane, then we need to know the singularities of the function that we want to integrate and their residues. We usually apply this technique to functions that we can express as a Laurent series,

(Eq=n 1)

for which z=x+iy and z0 representing a point at which the function has a singularity. We call the b-part of that double series the principle part of the Laurent representation and we can use its behavior near z0 to distinguish three types of singularity.

An analytic function of a complex variable has a singular point at z0 if it is not differentiable at z0 but has well-defined derivatives in all of the neighborhoods around z0. More precisely, we have an isolated singular point where the function f(z) has no derivatives at z0 but has well-defined derivatives at every point arbitrarily close to z0. We have three different kinds of isolated singularity, corresponding to the three ways in which the principle part of the Laurent series can Ablow up@ :

1. The removable singular point: If bn=0 for all values of n\$ 1, then z0 represents a removable singular point of the function f(z). In that case the Laurent series contains only positive powers of z-z0 and the singularity is purely an illusion. We have a function that looks like it involves a division by zero (which would definitely produce a singularity), but when we simplify the function or recast in a different representation we see that there is no actual division by zero. Consider for example

(Eq=n 2)

Our first impression of that function, gaining from looking at the formula on the left side of the equality sign, tells us that the function looks like it has a singularity at the point z=0, but when we recast the formula by bringing the infinite-series expansion of the exponential into play we find that the function does not actually involve a division by zero. Every example that I have seen of a removable singularity has that same illusory character.

2. The Pole: If bn=0 for all n>m and if bm0, then the function has a pole of order m at z0. If m=1, we call the singularity a simple pole. For example, we have

(Eq=n 3)

That function has a 3rd-order pole at z=0 and, as we shall soon be able to calculate, a residue of -1/2. Poles give us the most important type of singularity in applications to theories of physics.

3. The Essential Singularity: If the coefficients bn0 comprise an infinite set, then the function has an essential singularity at z0. The simplest example of such a singular point is the point z=0 for the function

(Eq=n 4)

for which the coefficients b=1/n! comprise an infinite set.

We can evaluate integrals of functions that have singularities by exploiting Cauchy=s integral theorem in the form of the residue theorem. If we have a function f(z) in a region of the complex plane enclosed within a closed curve C that we traverse in the positive (counterclockwise) direction and if that function has singularities in that region at the points zj, then

(Eq=n 5)

that is, the integral of the function on the curve equals 2πi multiplied by the sum of the residues of the singularities enclosed within the curve.

If within the contour C the function has a simple pole at z0, we can invert that equation and solve it for the residue; that is,

(Eq=n 6)

If f(z) has an m-th order pole, then its series expansion has a many as m terms that run toward infinity as z approaches z0. But all of those terms except the term b1/(z-z0) drop out of the calculation when we apply Cauchy=s integral formula. However, we don=t always get our function in a neat series that lets us pick off the residue

(Eq=n 7)

by inspection. In that case if we know the degree of the pole, then we know that the series expansion has no terms beyond bm/(z-z0)m, so we can calculate the residue by the formula

(Eq=n 8)

In essence we use the same technique that we use to calculate the m-th coefficient of a Taylor series, the multiplication of the raw function by the m-th power of (z-z0) compensating for the fact that our function has terms with negative powers of (z-z0), which would otherwise not zero out in the limit. We take the limit rather than use direct substitution of z0 in order to avoid obtaining an indeterminate form.

To give ourselves an example that illustrates these facts, let=s evaluate the integral

(Eq=n 9)

Because the integrand is an even function, f(-x)=f(x), we can extend the range of the integration onto the negative x-axis and then take our answer as half of the result,

(Eq=n 10)

In that way we push the part of our contour of integration that goes off the x-axis an infinite distance from the origin of the complex plane and thereby make its contribution to the integral equal to zero. Next we extend our integration onto the whole complex plane by replacing x with the complex variable z,

(Eq=n 11)

That integrand has two pairs of poles that we must take into account through their residues. The equation z2+1=0 tells us that the function in the integrand has simple poles at z=+i and at z=-i. The equation (z2+4)2=0 tells us that the function has 2nd-order poles at z=+2i and at z=-2i.

Now we must choose how we draw the contour around which we want to carry out the integration. We must certainly include the x-axis, proceeding from -4 to +4 . Then we can draw a curve that encloses the upper half of the complex plane (all y>0), in the limit, of course, because the complex plane has no outer edge, no actual infinity. In that case we integrate in the counterclockwise direction and obtain a positive result. Alternatively, we can draw a curve that encloses, in the limit as R goes toward infinity, the lower half of the complex plane (all y<0). In that case we carry out our contour integration in the clockwise direction and obtain a negative result, which we must then multiply by minus one to get the correct answer.

We choose to use the contour that encloses the upper half of the complex plane and now we want to calculate the residues of the two poles that lie in that realm. For the simple pole at z=+i we have

(Eq=n 12)

For the 2nd-order pole at z=+2i we have

(Eq=n 13)

So we get

(Eq=n 14)

And thus we find yet another example of the piety that we express through contour integration.

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