Method of Residues. Residue theorem. Evaluation of real definite integrals. Cauchy principal value. Summation of series.

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Method of Residues. Let f(z) be analytic in a region R, except for a singular point at z = a, as shown in Fig. 1. Cauchy’s theorem tells us that the integral of f(z) around any simple closed curve that doesn’t enclose any singular points is zero. Thus for a curve such as C₁ in the figure

What is the value of the integral of f(z) around a curve such as C₂ in the figure that does enclose a singular point? The answer:

where a_-1 is the coefficient of 1/(z - a) in the Laurent expansion of f(z) about a. The Laurent expansion about z = a is given by

where

If we evaluate 3) for k = -1 we get

The same result can be obtained by taking the integral of f(z) in 2)

and integrating term by term using the following theorem

Theorem 1. Let C be a simple closed curve containing point a in its interior. Then

Proof

whereby all terms except the a_-1 term drop out.

Def. Residue of an analytic function at an isolated singular point. The residue of an analytic function f(z) at an isolated singular point z₀ is

where a_-1 is the coefficient of (z - a)^-1 in the Laurent expansion of f(z) about z₀ and C is a simple closed curve enclosing z₀.

Residue theorem. Let f(z) be analytic inside and on a simple closed curve C except at the isolated singularities a, b, c, ... inside C which have residues given by a_r, b_r, c_r ... . Then

See Fig.2.

Proof

Calculation of residues

1. Residues at poles. The following theorem gives a simple procedure for the calculation of residues at poles.

Theorem 2. If f(z) has a pole of order m at z = a, then the residue of f(z) at z = a is given by

if m =1, and by

if m > 1.

Proof

Note. Formula 6) can be considered a special case of 7) if we define 0! = 1.

Example. Let

Then f(z) has two poles: z = -2, a pole of order 1, and z = 3, a pole of order 2.

The residue at z = -2 is given by

The residue at z = 3 is given by

Often the order of the pole will not be known in advance. In this case it is still possible to apply Theorem 2 by taking m = 1, 2, 3, ... , in turn, until the first time a finite limit is obtained for a_-1. The value of m for which this occurs is the order of the pole and the value of a_-1 thus computed is the residue.

2. Residues at essential points. Residues at essential singularities can sometimes be found by using known series expansions.

Example. In the function f(z) = e^-1/z, z = 0 is an essential singularity. Using the known series expansion for e^u

and setting u = -1/z we get the series expansion for e^-1/z

The residue at z = 0 is the coefficient of 1/z and is -1.

Theorem 3. The power series expansion of a function about a point is unique.

The Laurent expansion about a point is unique. Thus if a series expansion of the Laurent type is found by any process, it must be the Laurent expansion.

3. Residues at removable singularities.

Theorem 4. The residue of a function at a removable singularity is zero.

Evaluation of real definite integrals. There are several large and important classes of real definite integrals that can be evaluated by the Method of Residues. We now treat the following types:

Type 1. The integral

where the integrand R₁ is a finite-valued rational function of sin θ and cos θ for 0 θ 2π.

General procedure. Perform the substitution z = e^iθ. This substitution transforms integral 8) into the integral

where R₂(z) is a rational function of z and C is the positively-sensed unit circle centered at z = 0 shown in Fig. 3. The residue theorem then gives the solution of 9) as

where Σ r is the sum of the residues of R₂(z) at those singularities of R₂(z) that lie inside C.

Details. We perform the substitution z = e^iθ as follows: Apply the substitution to

thus transforming them into

and then substitute these expressions for sin θ and cos θ as expressed in terms of z and z^-1 into R₁(sin θ, cos θ). Let f(z) be the function obtained from R₁(sin θ, cos θ) by the substitution. From z = e^iθ we get dθ = dz/iz. Then R₂(z) = f(z)/iz. We can then calculate the residues of those singular points of R₂(z) that lie within the unit circle by methods described above and the integral is evaluated as

Type 2. The integral

where the integrand R(x) = P(x)/Q(x) is a rational function that has no poles on the real axis and is such that the degree of the polynomial Q(x) in the denominator is at least two greater than the degree of the polynomial P(x) in the numerator.

Theorem 5. If U(z) is a function which is analytic in the upper half of the z plane except at a finite number of poles, none of which are on the real axis, and if zU(z) converges uniformly to zero when z through values for which 0 arg z π, then is equal to 2πi times the sum of the residues at the poles of U(z) which lie in the upper half plane.

Proof

A function R(x) = P(x)/Q(x) automatically satisfies all the requirements of Theorem 5 if the degree of the denominator exceeds the degree of the numerator by at least two. Thus we have the following corollary.

Corollary 1. If P(x) and Q(x) are real polynomials such that the degree of Q(x) is at least two more than the degree of P(x), and if Q(x) has no real roots, then

General procedure. We determine the poles from the zeros of Q(x) and then compute the residues at the poles in the upper half plane by the method of Theorem 2 above.

Example. Evaluate the integral

Solution. The integral meets the requirements of Corollary 1. The only poles are at z = ai, bi. Only the poles ai and bi lie in the upper half plane. At z = ai the residue is

From symmetry it can be seen that the residue at z = bi must be b/2i(b² - a²). Thus the value of the integral is

Type 3. The integral

where the function R(x) = P(x)/Q(x) is a rational function that has no poles on the real axis and the degree of the polynomial in the denominator is at least one greater than that of the polynomial in the numerator.

Solution. Let

and consider the function R(z)e^imz . Let Σ r be the sum of the residues of R(z)e^imz in the upper half plane. Then

and I₁ and I₂ are, respectively, the real and imaginary parts of I. In other words,

Example. Evaluate

Solution. Consider the related function

Its only pole in the upper half plane is z = i, and its residue there is

Before proceeding to the next type we need to define the term Cauchy principal value.

The Cauchy principal value of integrals. Let f(x) be a function which is finite at all points of the closed interval [a, b] except at the point x = c, where it becomes infinite. See Fig. 4. Suppose that f(x) is integrable on the intervals [a, c - ε₁] and [c + ε₂, b] for any positive values of ε₁ and ε_2. Then we define

In some cases the above limit does not exist for ε₁ ε₂ but does exist if we take ε₁ = ε₂ = ε. In such a case we define

and call it the Cauchy principal value, or simply principal value, of integral and write

Example. The following integral

does not exist, however the Cauchy principal value with ε₁ = ε₂ = ε does exist and equals zero.

Type 4. The integral

where the associated complex function f(z) is a meromorphic function which may have simple poles on the real axis and which approaches zero uniformly on any circular arc centered at z = 0 as the radius of the arc approaches infinity. See Fig. 5.

Let a function f(z) satisfy the inequality |f(z)| < K_ρ when z is on a circular arc Γ_ρ of radius ρ, and let K_ρ depend only on ρ so that the inequality holds for all z on Γ_ρ, regardless of the argument of z. If K_ρ→ 0 as ρ → ∞, then f(z) approaches zero uniformly on Γ_ρ as ρ → ∞.

If ρ is allowed to become sufficiently large all poles in the upper half plane will fall within the contour shown in Fig. 5.

Solution. Consider the associated function f(z)e^imz = f(z) cos mx + f(z) sin mx. Let Σ r be the sum of the residues of f(z)e^imz at all poles lying in the upper half plane (not including those on the real axis). Let Σ r' be the sum of the residues of f(z)e^imz at all simple poles lying on the real axis. Then

Type 5. The integral

where Q(z) is analytic everywhere in the z plane except at a finite number of poles, none of which lies on the positive half of the real axis.

Solution. The solution is given by the following theorem:

Theorem. Let Q(z) be analytic everywhere in the z plane except at a finite number of poles, none of which lies on the positive half of the real axis. If |z^aQ(z)| converges uniformly to zero when z → 0 and when z → ∞, then

provided arg z is taken in the interval (-π, π).

It should be noted that unless a is an integer, (-z)^a-1 is a multiple-valued function which, using the defining formula a^z = e ^{z ln a}, is given by

(-z)^a-1 = e ^{(a-1) ln (-z)} = e ^{(a-1)[ln |z| + i arg (-z)}} -π < arg z π

Type 6. The integral

where R(z) is a rational function of z which has no poles at z = 0 nor on the positive part of the real axis and k is not an integer.

Solution. To insure convergence of this integral it is necessary that it have the proper behavior at zero and infinity. It is sufficient that . In evaluating the integral we employ the related function z^-kR(z) which is a multiple-valued function. The branch of this function that is used is z^-k = e ^{-k(ln |z| + i arg z)}.

Let Σ r be the sum of the residues of z^-kR(z) at the poles of R(z). The integral is evaluated by the formula

provided z^-k = e ^{-k(ln |z| + i arg z)}.

For types of integrals not covered above, evaluation by the method of residues, when possible at all, usually requires considerable ingenuity in selecting the appropriate contour and in eliminating the integrals over all but the selected portion of the contour.

Special theorems used in evaluating definite integrals. In evaluating definite integrals by the method of residues the following theorems are often useful.

Theorem 1. Let Γ_ρ be a semicircular arc of radius ρ, in the upper half plane, centered at the origin. See Fig. 6. If |f(z)| M/ρ^k for z = ρe^iθ where k > 1 and M are constants then

Proof

Theorem 2. Let Γ_ρ be a semicircular arc of radius ρ, in the upper half plane, centered at the origin. See Fig. 6. If |f(z)| M/ρ^k for z = ρe^iθ where k > 1 and M are constants then

Theorem 3. Let Γ_ρ be any circular arc of radius ρ centered at the origin. Let R(z) = P(z)/Q(z) be a rational function in which P(z) and Q(z) are polynomials and the degree of Q(z) is at least two greater than that of P(z). Then

Jordan’s lemma. Let Γ_ρ be a semicircular arc of radius ρ, in the upper half plane, centered at the origin. See Fig. 6. Let R(z) = P(z)/Q(z) be a rational function in which P(z) and Q(z) are polynomials and the degree of Q(z) is at least one greater than that of P(z). Then

Leibnitz’s rule for differentiation under the integral sign. A method sometimes useful for evaluating integrals utilizes Leibnitz’s rule for differentiation under the integral sign. The rule states that

The rule is valid if a and b are constants, α is a real parameter such that α₁ α α₂ where α₁ and α₁ are constants, and f(x, α) is continuous and has a continuous partial derivative with respect to α for a x b, α₁ α α₂. It can be extended to cases where the limits a and b are infinite or dependent on α.

Residue theorem used to sum series. . The residue theorem can often be used to sum various types of series. Let us denote an infinite series such as, for example,

by the notation

The following results are valid under very mild restrictions on f(z) which are usually satisfied whenever the series converge. Note that we replace n by the complex number z in the formula, viewing f(z) as complex.