4.5 The Residue Theorem
After Riemann and Weierstrass refined the understanding of analytic functions and the formal characterization of Jordan curves, the Cauchy Residue Theorem was consequently formalized. Cauchy had the informal notion of a residue, which we will now formally introduce.
Definition 4.5.1 (Residue).
For some , , suppose is holomorphic. Then the residue of at , denoted by or , is equal to
where is arbitrary. Since has a Laurent expansion at , being
we get that the residue of at is equal to the first term of the principal part of its Laurent expansion.
It then follows that the residue at a removable singularity is 0. As a direct consequence of (4.5.1), we can derive explicit formulas for the calculation of residues at poles. If is open, is an isolated singularity (a pole of order ) of that is holomorphic, then locally:
Multiplying by , we obtain that
By the definition of a Taylor series, we find that
Let be an isolated singularity of , which is holomorphic in , for sufficiently large finite . Then for finite , the residue at is defined as (notice the orientation)
Let . Then we get that
In this definition, if
the residue at is equal to . We will later explain the reasoning behind this definition.
Theorem 4.5.1 (Residue Theorem).
Let be an open set with a simple closed boundary curve . Suppose is a finite set and is holomorphic on and continuous on . Then,
Proof.
Since is open, there exists a small disk centered at each isolated singularity of radii . By the Cauchy–Goursat Theorem (Theorem 3.1.7), we get that
From rearrangement, , and the conclusion follows.
This result itself is fairly trivial. Now we will explain the significance of the residue at infinity.
Theorem 4.5.2 (Global Residue Theorem).
If is discrete and finite, and is holomorphic, and these points are the isolated singularities of , then the sum of the residues at each of these isolated singularities is zero, or
Proof.
Let be arbitrary. By the Residue Theorem (Theorem 4.5.1),
as desired. This is merely a restatement of Theorem 4.5.1.
There is not a directly trivial reason for the definition of the residue at , except for the fact that it seemingly “unifies” the Riemann sphere.
However, if we take a neighborhood of an arbitrary point in on the Riemann sphere and traverse its boundary clockwise (from the perspective of outside the sphere), its projection onto will be counterclockwise (Figure 7). However, the boundary of a neighborhood of in will have a clockwise projection (hence the difference in orientation). We define its equality with the residue of at , rather than , because we compose the differential form with the inversion, as opposed to .
For any closed rectifiable curve (here we are not bound under the assumption of simpleness), the Residue Theorem can be generalized into:
where are the singularities of in and is the winding index.
Residues are extremely important as they allow for simple evaluation of definite (most commonly improper) real-valued integrals. This is because oftentimes, residues at poles are generally easy to calculate and have an integral representation. We can integrate over a contour (a smooth closed curve) that contains the important part of the real interval. Oftentimes this is the most non-trivial step.
Example 4.5.1.
Evaluate the improper integral , where .
Proof.
Consider to be a closed semicircle with radius as in Figure 8. Notice that the function has singularities at only and , both of which are poles of order . By (4.5.2), the residue at is
The singularity at is not relevant, as it is not enclosed by the contour. By the Residue Theorem (Theorem 4.5.1), we have
We will now show that the integral over the semicircle vanishes as . Under the assumption that , since
which is integrable over , and we can commute the limit with the integral. Therefore, we have
Example 4.5.2 (Dirichlet Integral).
Evaluate the integral .
Proof.
It is common to use integration with parameters to approach this integral. However, we will now provide a solution via contour integration.
Let . Consider a closed contour in the form of Figure 9, consisting of a semicircle of radius in (), a line segment from to , a smaller semicircle of radius in the upper half-plane (), and a line segment from to .
By the Cauchy–Goursat Theorem (Theorem 3.1.7), we have that
We will now analyze each integral. The first integral is
Notice that over the integration range. We want to observe the behavior as :
Let us evaluate the integral on as :
Obviously,
and therefore, the integral and the limit may commute:
Evaluating the integral over the line segments, we have
Hence,
Example 4.5.3 (Fresnel Integral).
Evaluate the improper integrals
Proof.
Let . Choose the wedge contour composed of
as in Figure 10. By the Cauchy–Goursat Theorem (Theorem 3.1.7), we have that
The third integral can be written as
Using the fact that on the integration range, it can be bounded as
As , this integral tends to 0. Let on . Then, we have
From (4.5.3), we have that
Since , we have
Since , we have
as desired.
Example 4.5.4.
Evaluate the integrals , where is a rational function of and that is continuous on .
Proof.
Let . Consequently, we have , , and , implying that . Therefore, by the Residue Theorem (Theorem 4.5.1), letting , we have
where where are the isolated singularities of in .
Example 4.5.5.
Evaluate , where .
Proof.
Let and let in the principal branches of and . Then except for at the zeros of , is holomorphic.
The solutions to are . Choose an indented wedge contour (as there is a logarithmic branch point singularity at the origin) with an angle of (as in Figure 10). The only singularity it encloses is . Since it is a simple zero of , this singularity is a simple pole.
The contour is the union of the following curves:
where and . By the Residue Theorem (Theorem 4.5.1), we get that
By (4.5.2), it follows that
We can write the integral on in terms of :
We also have
It can also be shown that the integral is bounded by a vanishing function as :
Similarly, as ,
By letting and , we have
It follows that
Example 4.5.6.
Prove that the Fourier transform of is itself, or that
Proof.
Fix and let . Its poles in occur when , or equivalently, when , where .
Since
we have that is a constant multiple of . In particular, . Therefore, we can use a rectangular contour as shown in Figure 12. Let the sides be denoted by
The only enclosed singularity is a simple pole at (simple by evaluation of the Taylor expansion of the denominator). By the Residue Theorem (Theorem 4.5.1), we get that
By (4.5.2), we have
The sum of the horizontal line integrals is equal to
As , we have . The remaining two integrals can be written as
They can be bounded with
and
Since the integrands are continuous and uniformly convergent to with respect to , we have
as . By rearrangement of (4.5.4),
or that
which proves the result.
Contour integration provides a powerful method for evaluating real improper integrals by leveraging the Residue Theorem (Theorem 4.5.1). The primary challenge often lies in constructing a suitable contour in the complex plane that encloses the relevant singularities of the integrand while ensuring that the contribution from the contributions from the remaining segments of the contour either vanishes or can be calculated with ease.
If the function is even and integrated on a domain such as , then the integral can be extended to the entire real axis. If decays sufficiently rapidly in the upper half plane , a semicircular contour is generally preferable, as illustrated in Figure 8. In the presence of singularities on the contour itself, we can insert arc indentations around them, as shown in Figure 9.
If is a constant multiple of (a type of quasiperiodicity) for some , it is a strong indication to use a rectangular contour. If is a constant multiple of for some , a wedge-shaped contour is an appropriate choice.
In the case that there are indentations along the contour, we have
Theorem 4.5.3.
Let and let . Suppose is a holomorphic function on with a simple pole at . Let and define be a counterclockwise-oriented, connected arc subtending an angle . Then,
Proof.
Parameterize with , where and . Then,
Since has a simple pole at , we can write a Laurent expansion around as
where is holomorphic in a neighborhood of and .
Then for ,
So,
Let . Since is continuous on the disk , it is bounded. Therefore, letting , we have
Therefore,
In the case that a branch point singularity is present on the contour, we may attempt to rewrite the function in a way such that the branch point is irrelevant. Otherwise, there are two types of “keyhole contours” that can be used to avoid the branch cut.
Example 4.5.7.
Evaluate .
Proof.
Notice that the integrand itself has branch points at coinciding with the poles from the denominator. We can rewrite the integral as
Let , where concretely,
and , and let , where the branch for is chosen to satisfy , such as the principal branch. The only singularity of in the upper half plane is a simple pole at . By the Residue Theorem (Theorem 4.5.1), we have
By (4.5.2), we have
Additionally, for , since as , by virtue of , it follows that .
Since and
by (4.5.5), we have .