INTRODUCTION

In this section we shall see how residue theory can be used to evaluate real integrals of the forms

(1)

(2)

(3)

where F in (1) and f in (2) and (3) are rational functions. For the rational function f(x) = p(x)/q(x) in (2) and (3), we will assume that the polynomials p and q have no common factors.

Integrals of the Form F(cos θ, sin θ) d θ

The basic idea here is to convert an integral of form (1) into a complex integral where the contour C is the unit circle centered at the origin. This contour can be parameterized by z = cos θ + i sin θ = e^iθ, 0 ≤ θ ≤ 2π. Using

we replace, in turn, dθ, cos θ, and sin θ by

(4)

The integral in (1) then becomes

where C is |z| = 1.

EXAMPLE 1 A Real Trigonometric Integral

Evaluate

SOLUTION

Using the substitutions in (4) and simplifying yield the contour integral

With the aid of the quadratic formula we can write

where z₀ = −2 − and z₁ = −2 + . Since only z₁ is inside the unit circle C, we have

Now z₁ is a pole of order 2 and so from (2) of Section 19.5,

Hence,

and finally ≡

Integrals of the Form f(x) dx

When f is continuous on (−∞, ∞), recall from calculus that the improper integral f(x) dx is defined in terms of two distinct limits:

(5)

If both limits in (5) exist, the integral is said to be convergent; if one or both of the limits fail to exist, the integral is divergent. In the event that we know (a priori) that an integral f(x) dx converges, we can evaluate it by means of a single limiting process:

(6)

It is important to note that the symmetric limit in (6) may exist even though the improper integral is divergent. For example, the integral x dx is divergent since lim_R→∞ x dx = lim_R→∞ R² = ∞. However, using (6), we obtain

(7)

The limit in (6) is called the Cauchy principal value of the integral and is written

In (7) we have shown that P.V. x dx = 0. To summarize, when an integral of form (2) converges, its Cauchy principal value is the same as the value of the integral. If the integral diverges, it may still possess a Cauchy principal value.

To evaluate an integral f(x) dx, where f(x) = P(x)/Q(x) is continuous on (−∞, ∞), by residue theory we replace x by the complex variable z and integrate the complex function f over a closed contour C that consists of the interval [−R, R] on the real axis and a semicircle C_R of radius large enough to enclose all the poles of f(z) = P(z)/Q(z) in the upper half-plane Re(z) > 0. See FIGURE 19.6.1. By Theorem 19.5.3 we have

where z_k, k = 1, 2, . . ., n, denotes poles in the upper half-plane. If we can show that the integral f(z) dz → 0 as R → ∞, then we have

(8)

EXAMPLE 2 Cauchy P.V. of an Improper Integral

Evaluate the Cauchy principal value of

SOLUTION

Let f(z) = 1/(z² + 1)(z² + 9). Since

(z² + 1)(z² + 9) = (z − i)(z + i)(z − 3i)(z + 3i),

we let C be the closed contour consisting of the interval [−R, R] on the x-axis and the semicircle C_R of radius R > 3. As seen from FIGURE 19.6.2,

and I₁ + I₂ = 2πi[Res (f(z), i) + Res (f(z), 3i)].

At the simple poles z = i and z = 3i we find, respectively,

so that (9)

We now want to let R → ∞ in (9). Before doing this, we note that on C_R,

|(z² + 1)(z² + 9)| = |z² + 1| |z² + 9| ≥ | |z|² − 1| | |z|² − 9| = (R² − 1)(R² − 9),

and so from the ML-inequality of Section 18.1 we can write

This last result shows that |I₂| → 0 as R → ∞, and so we conclude that lim_R→∞ I₂ = 0. It follows from (9) that lim_R→∞ I₁ = π/12; in other words,

≡

It is often tedious to have to show that the contour integral along C_R approaches zero as R → ∞. Sufficient conditions under which this is always true are given in the next theorem.

In other words, the integral along C_R approaches zero as R → ∞ when the denominator of f is of a power at least 2 more than its numerator. The proof of this fact follows in the same manner as in Example 2. Notice in that example that the conditions stipulated in Theorem 19.6.1 are satisfied, since the degree of P(z) = 1 is 0 and the degree of Q(z) = (z² + 1)(z² + 9) is 4.

EXAMPLE 3 Cauchy P.V. of an Improper Integral

Evaluate the Cauchy principal value of

SOLUTION

By inspection of the integrand, we see that the conditions given in Theorem 19.6.1 are satisfied. Moreover, we know from Example 3 of Section 19.5 that f has simple poles in the upper half-plane at z₁ = e^πi/4 and z₂ = e^3πi/4. We also saw in that example that the residues at these poles are

Thus, by (8),

≡

Integrals of the Forms f(x) cos αx dx or f(x) sin αx dx

We encountered integrals of this type in Section 15.4 in the study of Fourier transforms. Accordingly, f(x) cos αx dx and f(x) sin αx dx, α > 0, are referred to as Fourier integrals. Fourier integrals appear as the real and imaginary parts in the improper integral f(x)e^iαx dx. Using Euler’s formula e^iαx = cos αx + i sin αx, we get

(10)

whenever both integrals on the right side converge. When f(x) = P(x)/Q(x) is continuous on (−∞, ∞) we can evaluate both Fourier integrals at the same time by considering the integral , where α > 0 and the contour C again consists of the interval [−R, R] on the real axis and a semicircular contour C_R with radius large enough to enclose the poles of f(z) in the upper half-plane.

Before proceeding we give, without proof, sufficient conditions under which the contour integral along C_R approaches zero as R → ∞:

EXAMPLE 4 Using Symmetry

Evaluate the Cauchy principal value of .

SOLUTION

First note that the limits of integration are not from −∞ to ∞ as required by the method. This can be rectified by observing that since the integrand is an even function of x, we can write

(11)

With α = 1, we now form the contour integral

where C is the same contour shown in Figure 19.6.2. By Theorem 19.5.3,

where f(z) = z/(z² + 9). From (4) of Section 19.5,

Hence, in view of Theorem 19.6.2 we conclude f(z) e^iz dz → 0 as R → ∞ and so

But by (10),

Equating real and imaginary parts in the last line gives the bonus result

Finally, in view of (11) we obtain the value of the prescribed integral:

≡

Indented Contours

The improper integrals of form (2) and (3) that we have considered up to this point were continuous on the interval (−∞, ∞). In other words, the complex function f(z) = P(z)/Q(z) did not have poles on the real axis. In the event f has poles on the real axis, we must modify the procedure used in Examples 2–4. For example, to evaluate f(x) dx by residues when f(z) has a pole at z = c, where c is a real number, we use an indented contour as illustrated in FIGURE 19.6.3. The symbol C_r denotes a semicircular contour centered at z = c oriented in the positive direction. The next theorem is important to this discussion.

Suppose f has a simple pole z = c on the real axis. If C_r is the contour defined by z = c + re^iθ, 0 ≤ θ ≤ π, then

PROOF:

Since f has a simple pole at z = c, its Laurent series is

where a₋₁ = Res (f(z), c) and g is analytic at c. Using the Laurent series and the parameterization of C_r , we have

(12)

First, we see that

Next, g is analytic at c and so it is continuous at this point and bounded in a neighborhood of the point; that is, there exists an M > 0 for which |g(c + re^iθ)| ≤ M. Hence,

It follows from this last inequality that lim_r→0|I₂| = 0 and consequently lim_r→0 I₂ = 0. By taking the limit of (12) as r → 0, we have proved the theorem. ≡

EXAMPLE 5 Using an Indented Contour

Evaluate the Cauchy principal value of

SOLUTION

Since the integral is of form (3), we consider the contour integral e^iz dz/z(z² − 2z + 2). The function f(z) = 1/z(z² − 2z + 2) has simple poles at z = 0 and at z = 1 + i in the upper half-plane. The contour C shown in FIGURE 19.6.4 is indented at the origin. Adopting an obvious notation, we have

(13)

where Taking the limits of (13) as R → ∞ and as r → 0, we find from Theorems 19.6.2 and 19.6.3 that

Now,

Therefore,

Using e⁻¹⁺ⁱ = e⁻¹(cos 1 + i sin 1), simplifying, and then equating real and imaginary parts, we get from the last equality

and ≡

19.6 Exercises Answers to selected odd-numbered problems begin on page ANS-48.

In Problems 1–10, evaluate the given trigonometric integral.

1. 
2. 
3. 
4. 
5. [Hint: Let t = 2π − θ.]
6. 
7. 
8. 
9. 
10. 

In Problems 11–30, evaluate the Cauchy principal value of the given improper integral.

In Problems 31 and 32, use an indented contour and residues to establish the given result.

31. 
32. 
33. Establish the general result

    

    and use this formula to verify the answer in Example 1.

34. Establish the general result

    

    and use this formula to verify the answer to Problem 7.

35. Use the contour shown in FIGURE 19.6.5 to show that

    

    

36. The steady-state temperature u(x, y) in a semi-infinite plate is determined from

    

    Use a Fourier transform and the residue method to show that