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

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 C_{1} in the
figure

What is the value of the integral of f(z)
around a curve such as C_{2} 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

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 z_{0} is

where a_{-1} is the coefficient of (z - a)^{-1} in the
Laurent expansion of f(z) about z_{0} and C is a
simple closed curve enclosing z_{0}.

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.

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.

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 R_{1} 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 R_{2}(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 R_{2}(z) at those singularities of R_{2}(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
R_{1}(sin θ, cos θ). Let f(z) be the function obtained from R_{1}(sin θ, cos θ) by the substitution. From
z = e^{iθ} we get dθ = dz/iz. Then R_{2}(z) = f(z)/iz. We can then calculate the residues of those
singular points of R_{2}(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.

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(b^{2} - a^{2}). 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 I_{1} and I_{2} 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

So

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 - ε_{1}] and [c + ε_{2}, b] for any positive values of
ε_{1} and ε_{2. } Then we define

In some cases the above limit does not exist for ε_{1}
ε_{2} but does exist if we take ε_{1} = ε_{2} = 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 ε_{1} = ε_{2} = 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^{a}Q(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^{-k}R(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^{-k}R(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

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 α_{1}
α
α_{2} where α_{1} and
α_{1} are constants, and f(x, α) is continuous and has a continuous partial derivative with respect to
α for a
x
b, α_{1}
α
α_{2}. 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.

Example. Evaluate

where a > 0.

Solution. Let

which has simple poles at z = ±ai.

The residue of

at z = ai is

Similarly the residue at z = -ai is

The sum of the two residues is then

Consequently,

References

Spiegel. Complex Variables (Schaum)

Wylie. Advanced Engineering Mathematics

Hauser. Complex Variables with Physical Applications.

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