CHAPTER 6
Residue Theory
We developed three methods of evaluating contour integrals around closed curves: (1) The Cauchy-Goursat Theorem - but this requires the integrand to be ana- lytic in and on the curve; (2) Cauchy’s integral formulas - but only applies to integrands that have a very special form; (3) Theorem 4.8 - but replaces a contour integral along a curve containing a number of singularities with contour integrals along curves each of which encloses only one singularity. Cauchy’s residue theorem, developed in this chapter, encompasses all of these. 6.1. Cauchy’s Residue Theorem Cauchy’s residue theorem is a simple consequence of the equations for coe - cients of a Laurent series. 1 To illustrate, suppose f(z) = a (z z )n is the Laurent series of a function n 0 n= f valid in some annulus 0 < z X 1z < R about an isolated singularily z . When | 0| 0 C is a simple, closed, piecewise smooth curve in this annulus that contains z0 in its interior, coe cient a 1 is defined by
146 6.1. CAUCHY’S RESIDUE THEOREM 147
1 a 1 = f(z) dz = f(z)dz = 2⇡ia 1. 2⇡i C ) c Cauchy’s residue theorem Iis a simple extensionI of this result to the case when C encloses an arbitrary number of isolated singularities. Definition (6.1). The residue of a function f at an isolated singularity z0 is denoted by Res[f, z0]. It is the coe cient a 1 in the Laurent expansion of f about z0. From the examples on pages 141-142 of these notes, sin z Res , 0 = 0, Res[e1/z, 0] = 1 z h z i sin z 1 Res , 0 = z6 120 Theorem (6.1 – Cauchy’sh Residue Theorem)i . Suppose a function f is analytic inside and on a simple, closed, piecewise smooth curve C, except at singularies z1, . . . , zn in its interior. Then n
f(z) dz = 2⇡i Res[f, zj]. C j=1 I X Proof. Construct around the z disjoint circles C : z z = r such j j | j| j that each Cj is interior to C (see the diagram at the top of the next page). By Theorem 4.8, the contour integral around C can be replaced by contour integrals around the Cj, 148 6. RESIDUE THEORY
n f(z) dz = f(z) dz. C j=1 Cj I X I If f is expanded in a Laurent series about zj,
1 n 1 f(z) = an(z zj) where a 1 = f(z) dz = 2⇡i ) n= Cj X 1 I
f(z) dz = 2⇡ia 1 = 2⇡i Res[f, zj] = ) ICj n f(z) dz = 2⇡i Res[f, zj]. ⇤ C j=1 I X 1 Problem (Page 268 #13). Evaluate dz where C : z = 1. z2 + 4z + 16 | | IC Solution. 4 p16 64 z2 + 4z + 16 = 0 z = ± = 2 2p3 i. () 2 ± Both of these values are exterior to C, as seen in the diagram at the top of the next page. Thus 1 dz = 0. z2 + 4z + 16 IC ⇤ 6.1. CAUCHY’S RESIDUE THEOREM 149
1 Example. Evaluate sin dz where C : z = 2. C z | | I ⇣ ⌘ 1 Solution. Since the Laurent series for sin around z = 0 is z 1 1 1 1⇣ ⌘ sin = + , z z 3!z3 5!z5 · · · 1 ⇣ ⌘ 1 the residue of sin is Res sin , 0 = 1. Thus z z ⇣ ⌘ ⇣ ⌘ 1 sin dz = 2⇡i(1) = 2⇡i. z ⇤ IC The following theorem giv⇣es ⌘a formula for calculating residues at poles.
Theorem (6.2). If f has a pole of order m at z0, then m 1 1 d m Res[f, z0] = lim m 1 (z z0) f(z) z z0 (m 1)! dz ! Proof. f has a pole of order m at z = (Lauren⇥ t expansion)⇤ 0 ) a m a m+1 a 1 f(z) = + + + + a + , (z z )m (z z )m 1 · · · (z z ) 0 · · · 0 0 0 valid in some annulus 0 < z z < R. Thus | 0| m m 1 (z z0) f(z) = a m + a m+1(z z0) + + a 1(z z0) + = · · · · · · ) m 1 d m m 1 (z z0) f(z) = (m 1)!a 1 + m!a0(z z0) + = dz · · · ) ⇥ ⇤ 150 6. RESIDUE THEORY m 1 1 d m lim m 1 (z z0) f(z) = a 1 = Res[f, z0] ⇤ z z0 (m 1)! dz ! Problem (Page 268 # 1).⇥ Calculate the ⇤residue for the isolated singularity (z + 3)6 of at z = 4i. (z 4i)4 Solution. The isolated singularity at z = 4i is a pole of order 4 = ) (z + 3)6 1 d3 (z 4i)4(z + 3)6 Res , 4i = lim (z 4i)4 3! z 4i dz3 (z 4i)4 ! ⇢ 1 3 = lim 6(5)(4)(z + 3) = 2340 + 880i. ⇤ 6 z 4i ! ⇥ 1 ⇤ Problem (Page 268 # 21). Evaluate dz where C : z = 3. z2 z + 2 | | IC 1 1 Solution. = has simple poles at z = 2 and z2 z + 2 (z + 2)(z 1) z = 1, both interior to C.
1 z 1 1 Res , 1 = lim = , and z2 z + 2 z 1 (z + 2)(z 1) 3 ! 1 z + 2 1 Res , 2 = lim = = z2 z + 2 z 2 (z + 2)(z 1) 3 ) ! 1 1 1 dz = 2⇡i = 0. z2 z + 2 3 3 ⇤ IC ✓ ◆ 6.1. CAUCHY’S RESIDUE THEOREM 151 An important conequence of the residue theorem is the principle of the ar- gument. Theorem (6.3 – Principle of the Argument). Let f be analytic in a do- main D except at a finite number of poles. If C is a simple, closed, piece- wise smooth curve in D which does not pass through any poles or zeros of f, then f (z) ( ) 0 dz = 2⇡i(N P ) ⇤ f(z) IC where N and P are the sums of the orders of the zeroes and poles of f within C.
To see why this is called the principle of the argument,we transform the contour integral ( ) from the z-plane to the w-plane with w = f(z). It follows that ⇤ 1 f (z) 1 1 N P = 0 dz = dw, 2⇡i f(z) 2⇡i w IC If(C) where f(C) is the image of C in the w-plane. But according to text Example 4.22 (page 182), the contour integral on the right is the number of times that f(C) encircles w = 0. In other words, N P measures how many times the curve f(C) encircles w = 0 as z traverses C in the counterclockwise direction; that is N P measures how many times the argument of f(z) increases by 2⇡ as z traverses C. 152 6. RESIDUE THEORY Problem (Page 269 # 39). Use the principle of the argument to evaluate z sinh z + 3 cosh z dz where C : z i = 7. z cosh z | | IC Solution. Letting f(z) = z cosh z, we have z sinh z + 3 cosh z z sinh z + cosh z 2 dz = + dz z cosh z z cosh z z IC IC ✓ ◆ f (z) 1 = 0 dz + 2 dz = 2⇡i(N P ) + 2(2⇡i), f(z) z IC IC where N and P are the sums of the orders of the zeroes and poles of f(z) inside C.
There are 6 zeroes of order 1 and no poles within this circle, so N P = 6. Then z sinh z + 3 cosh z dz = 2⇡i(6) + 4⇡i = 16⇡i. z cosh z ⇤ IC 6.1. CAUCHY’S RESIDUE THEOREM 153 Theorem (6.4 – Rouche’s Theorem). Let fand g be analytic in a domain D. If C us a simple, closed, piecewise smooth curve in D which contains in its interior only points of D, and if g(z) < f(z) on C, then sums of the orders of the zeros of f + g and f inside| | C |are the| same. Proof. g < f on C = f(z) = 0 and f(z)+g(z) f(z) g(z) > 0 | | | | f)(z) + g6(z) | | | | | | on C. The function F (z) = , then, does not have any zeros or poles f(z) on C. Applying the principle of the argument to F and C, 1 F (z) 0 dz = N P, 2⇡i F (z) IC and each side may be interpreted as the number of times w = F (z) encircles w = 0 as z traverses C. But on C, f(z) + g(z) g(z) F (z) 1 = 1 = < 1 = | | f(z) f(z) ) the curve F (C) cannot encircle w = 0, and th us N = P .
Since the zeros of F occur at the zeros of f + g and the poles of F occur at the zeros of f, it follows that the sum of the orders of the zeros of f and f + g inside C must be the same. ⇤ 154 6. RESIDUE THEORY Problem (Page 269 # 45). Show that if g is analytic inside and on C : z = 1 and if g(z) < 1, then there is exactly one point z inside C for which g| (|z) = z. | | Solution. Let f(z) = z. Then, on C, f(z) = 1. Since g(z) < f(z) on C, it follows from Rouche’s theorem that|f(z)|+ g(z) = g(|z) |z has| the| same number of zeros inside C as f(z) = z, namely one. Thus, g(z) z has a single zero inside C = there is exactly one solution for g(z) = z inside ) C. ⇤ 6.2. Evaluation of Definite Intrgrals Contour integrals and residues can be used to evaluate certain classes of definite integrals that are otherwise di cult or impossible. Definite Integrals Involving Trigonometric Functions Contour integrals and residues can be useful in the evaluation of definite inte- grals that are rational functions of sines and cosines, b P (cos ✓, sin ✓) d✓ Q(cos ✓, sin ✓) Za where P (cos ✓, sin ✓) and Q(cos ✓, sin ✓) are polynomials in cos ✓ and sin ✓ pro- vided Q is never equal to 0 in the interval [a, b]. The technique is to set z = ei✓. This transforms the definite integral to a contour integral along an arc of the unit circle in the complex z-plane. When a = 0 and b = 2⇡, it is the complete circle, and residues are used to evaluate the contour integral. For other intervals of integration, residues may not be applicable, but other techniques for evaluating contour integrals may be advantageous. The integrand for the contour integral is obtained by substituting ✓i ✓i ✓ih ✓i e + e 1 1 e e 1 1 cos ✓ = = z + , sin ✓ = = z . 2 2 z 2i 2i z ✓ ◆ ✓ ◆ 6.2. EVALUATION OF DEFINITE INTRGRALS 155 2⇡ 1 Problem (Page 280 # 3). Evaluate d✓. 6 + 5 sin ✓ Z0 Solution. Set z = e✓i = dz = ie✓id✓ = ) ) 2⇡ 1 1 dz 2 d✓ = = 2 dz, 0 6 + 5 sin ✓ C z 1/z iz C 5z + 12iz 5 Z I 6 + 5 2i I where C is the circle z = 1. Since ⇣ ⌘ | | 12i p 144 + 100 ( 6 p11)i 5z2 + 12iz 5 = 0 = z = ± = ± ) 10 5 ( 6 + p11)i with only z = interior to C. Then 5
2 6 + p11 Res , i 5z2 + 12iz 5 5 ✓ ◆ 2 z ( 6 + p11)i/5 = lim z ( 6+p11)i/5 5 z ( 6 + p11)i/5 z ( 6 p11)i/5 ! ⇥ ⇤ 2 i ⇥= ⇤⇥ =⇤ . 5 ( 6 + p11)i/5 ( 6 p11)i/5 p11 Thus ⇥ ⇤ 2⇡ 1 2 i 2⇡ d✓ = dz = 2⇡i = . ⇤ 6 + 5 sin ✓ 5z2 + 12iz 5 p11 p11 Z0 IC ✓ ◆ 156 6. RESIDUE THEORY 0 1 Problem (Page 280 # 15). Evaluate d✓. ⇡ 4 + cos ✓ Z Solution. Set z = e✓i = dz = ie✓id✓ = ) ) 0 1 1 dz 1 d✓ = z+1/z = 2i 2 ⇡ 4 + cos ✓ C 4 + iz C z + 8z + 1 Z Z 2 Z where C is the semicircle shown below.
8 p64 4 z2 + 8z + 1 = 0 = z = ± = 4 p15. ) 2 ± Using partial fractions, 0 1 p15/30 p15/30 d✓ = 2i dz ⇡ 4 + cos ✓ C z + 4 p15 z + 4 + p15 Z Z ✓ ◆ i 1 = log⇡/2(z + 4 + p15) log⇡/2(z + 4 p15) p 1 15 i n o = log (5+p15) log (5 p15) log (3+p15)+log (3 p15) p15 ⇡/2 ⇡/2 ⇡/2 ⇡/2 i h i = ln(5+p15)+2⇡i ln(5 p15) 2⇡i ln(3+p15) 2⇡i+[ln(p15 3)+⇡i 15 h⇡ i 5 + p15 p15 3 ⇡ i 2p15 ⇡ i = + ln = + ln = . p15 p15 5 p15 · p15 + 3 p15 p15 2p15 p15 ✓ ◆ ⇤ 6.2. EVALUATION OF DEFINITE INTRGRALS 157 Real Improper Integrals Contour integrals are also e↵ective in evaluating improper integrals which have infinite upper and/or lower limits. Imprper Integrals of rational Functions The first type of improper integral we consider involve integrands that are rational functions, integrals of the form b P (x) dx, Q(x) Za where P (x) and Q(x) are polynomials, and at least one of a and b is infi- nite. They are simplest when Q(x) = 0 for any value of x in the interval of integration. 6
1 1 Example (6.8). Evaluate dx 1 + x4 Z 1 1 Solution. We want to evaluate the contour integral 4 dz where C C 1 + z is the contour shown below: I
If we let R , then the part of the contour integral along the real axis would give us the!required1 improper integral. We first find the contour integral over 4 1 C. The integrand (1 + z ) has simple poles at the four fourth roots of 1: e⇡i/4, e3⇡i/4, e5⇡i/4, e7⇡i/4. 158 6. RESIDUE THEORY As shown in the preceding diagram, only the first two are within C. Using L’Hopital’s rule (Theorem 5.24), ⇡i/4 1 ⇡i/4 z e 1 1 p2 Res 4 , e = lim 4 = lim 3 = 3⇡i/4 = (1 + i). 1 + z z e⇡i/4 1 + z z e⇡i/4 4z 4e 8 ! ! 1 p2 Similarly, Res , e3⇡i/4 = (1 i). By Cauchy’s residue theorem, 1 + z4 8 1 p2 p2 ⇡ dz = 2⇡i (1 + i) + (1 i) = . 1 + z4 8 8 p2 IC Supposxe we now divide C into a semicircular part and a straight line part as in the diagram below:
Then ⇡ R 1 1 = 4 dx + 4 dz. p2 R 1 + x 1 + z Z Z Set z = Re✓i, 0 ✓ ⇡ on = (by the triangle inequality) ) 1 1 1 = . 1 + z4 z4 1 R4 1 | | By the inequality at the top of page 96 of these notes, 1 1 1 1 ⇡ 4 dz 4 (⇡R) 0 as R = 4 dx = . ⇤ 1 + z R 1 ! ! 1 ) 1 + x p2 Z Z 1 The following theorem, based on the idea of the previous example, makes the evaluation of many examples easier. 6.2. EVALUATION OF DEFINITE INTRGRALS 159 Theorem (6.5). Suppose that P (x) and Q(x) are polynomials of degrees m and n, where n m + 2, and Q(x) = 0 for all x. Then 6 1 P (x) sum of the residues of P (z)/Q(z) at dx = 2⇡i Q(x) its poles in the half-plane Im z > 0 Z 1 ( 2 1 x + 3 Problem (Page 280 # 23). Evaluate dx. (x2 + 1)(x2 x + 1) Z 1 z2 + 3 Solution. The function has simple poles at z = i (z2 + 1)(z2 z + 1) ± 1 p3 i and z = ± , only two of which have positive imaginary parts. 2 z2 + 3 (z i)(z2 + 3) 2 Res , i = lim = = 1. (z2 + 1)(z2 z + 1) z i (z i)(z + i)(z2 z + 1) (2i)( i) ! z2 + 3 1 p3i Res , + (z2 + 1)(z2 z + 1) 2 2 1 p3i 2 z 2 2 (z + 3) = lim z 1/2+p3i/2 (z2 + 1)⇣ z 1 + p3i⌘ z 1 p3i ! 2 2 2 2 2 1 p3i ⇣ ⌘⇣ ⌘ 2 + 2 + 3 5 + p3 i 3 p3 i = = ⇣ 2 ⌘ p · p 1 p3i p 3 + 3 i 3 3 i 2 + 2 + 1 ( 3 i) ⇣ ⌘ 12 8p3 i 3 2p3 i = = . 12 3 Then by Theorem 6.5, 2 1 x + 3 2p3 i 4⇡p3 dx = 2⇡i 1 1 = . (x2 + 1)(x2 x + 1) 3 3 Z 1 ✓ ◆ ⇤ 160 6. RESIDUE THEORY 1 1 Example (6.10). Evaluate dx. 1 + x3 Z0 1 Solution. We consider dz where C is some appropriate contour. 1 + z3 IC The integrand has simple poles at e⇡i/3, 1,and e5⇡i/3. Only the first is in the upper half-plane, but we cannot use Theorem 6.5 since z = 1 is on the negative real axis. We choose the line z = re i, 0 < < ⇡.
⇡ Let’s choose such that < < ⇡, but, for the moment, leave it arbitrary. 3 Cauchy’s residue theorem and L’Hopital’s rule give ⇡i/3 1 1 ⇡i/3 z e 3 dz = 2⇡i Res 3 , e = 2⇡i lim 3 C 1 + z 1 + z z e⇡i/3 1 + z I ! 1 2⇡i (p3 i)⇡ = 2⇡i lim 2 = 2⇡i/3 = . z e⇡i/3 3z 3e 3 ! Thus (p3 i)⇡ 1 1 1 ( ) = dz + dz + dz. ⇤ 3 1 + z3 1 + z3 1 + z3 Z 1 Z 2 Z 3 ✓i Set z = Re on 1. On this arc, 1 1 1 1 1 = = dz (R ) 0 1 + z3 z 3 1 R3 1 ) 1 + z3 R3 1 ! Z 1 | | as R . Thus, letting R , ! 1 ! 1 6.2. EVALUATION OF DEFINITE INTRGRALS 161
(p3 i)⇡ 1 1 1 = 3 dz + 3 dx 3 2 1 + z 0 1 + x Z ⇡ Z where : z = re i, 0 r < , < < ⇡. Then 2 1 3 0 1 1 i i 1 1 3 dz = 3 3 i e dr = e 3 3 i dr 1 + z 1 + r e 0 1 + r e Z 2 Z1 Z 2⇡ Now we choose = = 3 ) 1 1 1 dz = e2⇡i/3 dr = 1 + z3 1 + r3 ) Z 2 Z0 (p3 i)⇡ 1 p3 i 1 1 = + + 1 dx = 3 2 2 1 + x3 ) ✓ ◆ Z0 1 1 (p3 i)⇡ 2 2⇡ 3 dx = = ⇤ 0 1 + x 3 · 3 p3 i 3p3 Z Improper Integrals of Rational Functions Multiplied by Sines and/or Cosines We now consider improper integrals of the forms
1 P (x) 1 P (x) cos ax dx and sin ax dx, Q(x) Q(x) Z 1 Z 1 where a > 0 is a constant. The following theorem accomplishes the same thing here that Theorem 6.5 did earlier for rational functions. 162 6. RESIDUE THEORY Theorem (6.6). Suppose that P (x) and Q(x) are polynomials of degrees m and n, where n m + 1, and Q(x) = 0 for all real x. When a > 0 is a constant, 6 azi 1 P (x) cos ax sum of the residues of P (z)e /Q(z) dx = 2⇡ Im Q(x) at its poles in the half-plane Im z > 0 Z 1 ( and azi 1 P (x) sin ax sum of the residues of P (z)e /Q(z) dx = 2⇡ Re . Q(x) at its poles in the half-plane Im z > 0 Z 1 ( 2 1 x cos x Problem (Page 280 # 33). Evaluate dx. (x2 + 9)2 Z 1 z2 cos z Solution. has double poles at 3i, but only 3i is in the upper (z2 + 9)2 ± half-plane. z2ezi d z2ezi(z 3i)2 d z2ezi Res , 3i = lim = lim (z2 + 9)2 z 3i dz (z + 9)2 z 3i dz (z + 3i)2 ! ! (z + 3i)2(2zezi + iz2ezi) z2ezi(2)(z + 3i) = lim z 3i (z + 3i)4 ! 3 3 3 6i 6ie + i( 9)e 2( 9)e i = = . h (6i)3 i 6e3 Then, by Theorem 6.6, 2 1 x cos x i ⇡ dx = 2⇡ Im = . (x2 + 9)2 6e3 3e3 Z 1 ✓ ◆ ⇤ 6.3. MORE IMPROPER INTEGRALS BY RESIDUES 163 6.3. More Improper Integrals by Residues The improper integral of a continuous function f(x) over the real line is defined by 0 b 1 ( ) f(x) dx = lim f(x) dx + lim f(x) dx ⇤ a a b 0 Z 1 ! 1 Z !1 Z provided both limits exist. The Cauchy principal value of the integral s defined by b 1 ( ) P V f(x) dx = lim f(x) dx ⇤⇤ b b Z 1 !1 Z provided the limit exists. ( ) = ( ) with the same value, but the converse is not universally true. ⇤ ) ⇤⇤ 1 x 1 x dx = 0 in both senses, but P V dx = 0, but does not x4 + 1 x2 + 1 conZ 1verge in the sense of ( ). Z 1 ⇤ Cauchy principal values are also defined for improper integrals due to discon- tinuities in integrands. When the only discontinuity of a function f(x) in the interval a < x < b is x = d, its improper integral over the interval a x b is defined by b d ✏ b (#) f(x) dx = lim f(x) dx + lim f(x) dx ✏ 0+ ✏ 0+ Za ! Za ! Zd+✏ provided both limits exist. The integral is said to have a Cauchy principal value when the limit b d ✏ b (##) P V f(x) dx = lim f(x) dx + f(x) dx ✏ 0+ Za ! Za Zd+✏ exists. Again, (#) = (##), but the converse is not uiversally true. ) 1 1 P V dx = 0, but has no value in the sense of (#). 1 x Z 164 6. RESIDUE THEORY Improper Integrals of Rational Functions with Simple Poles.
1 P (x) The first type of integral we consider is dx where P (x) and Q(x) are Q(x) polynomials, and Q(x) has real simple Zzeros. 1 Theorem (6.7). When P (x) and Q(x) are polynomials of degrees m and n where n m + 2, and the only zeros of Q(x) on the real axis are simple zeros, P (z) 1 P (x) sum of the residues of at P V dx = 2⇡i Q(z) Q(x) 8 Z 1 1 p3 i 2 2 z (z + 2z) z + 2z 1 + p3 2 2 Res 3 , = lim 3 (z 1)(z + 1) 2 z (1+p3 i)/2 ⇣ (z 1)(z⌘+ 1) ! z 1 p3 i (2z + 2) + (z2 + 2z) 2 2 = lim 3 2 z (1+p3 i)/2 ⇣ (z + 1)⌘+ (z 1)(3z ) ! z2 + 2z z + 2 = lim 2 = lim z (1+p3 i)/2 3z (z 1) z (1+p3 i)/2 3z(z 1) ! ! 1 + p3 i + 2 5 + p3 i = 2 2 = . 3 1 + p3 i 1 + p3 i 6 2 2 2 2 Then, by Theorem 6.7, ⇣ ⌘⇣ ⌘ 2 1 x + 2x 5 + p3 i 3 1 ⇡ P V dx = 2⇡i + ⇡i + = . ⇤ (x 1)(x3 + 1) 6 2 6 p3 Z 1 ✓ ◆ ✓ ◆ Improper Intergrals of Rational Functions with Simple Poles Multiplied by Sines and/or Cosines Theorem (6.8). When P (x)and Q(x) are polynomials of degrees m and n where n m + 1, and the only zeros of Q(x) are simple zeros, then for a > 0, P (z)eazi 1 P (x) cos ax sum of the residues of at P V dx = 2⇡ Im Q(z) Q(x) 8 Z 1 P (z)eazi 1 P (x) sin ax sum of the residues of at P V dx = 2⇡ Re Q(z) Q(x) 8 Z 1 1 sin x cos x Problem (Page 295 # 7). Ev:aluate P V dx. x3 8 Z 1 Solution. First, 1 sin x cos x 1 1 sin 2x P V dx = P V dx. x3 8 2 x3 8 Z 1 Z 1 z3 8 = (z 2)(z2 + 2z + 4) has simple poles at z = 2 and z = 1 p3 i. ± e2zi (z 2)e2zi e4i Res , 2 = lim = , z3 8 z 2 (z 2)(z2 + 2z + 4) 12 ! e2zi (z + 1 p3 i)e2zi p Res 3 , 1 + 3 i = lim 3 z 8 z 1+p3 i z 8 ! 2zi 2zi 2( 1+p3 i)i e + 2i(z + 1 p3 i)e e = lim 2 = z 1+p3 i 3z 3(1 + p3 i)2 ! 2(p3+1) 2(p3+i) e 1 p3 i (1 p3 i)e = = . 6(1 + p3 i) · 1 p3 i 24 By Theorem 6.8, 6.3. MORE IMPROPER INTEGRALS BY RESIDUES 167 1 sin x cos x 1 1 sin 2x P V dx = P V dx x3 8 2 x3 8 Z 1 Z 1 4i 1 2(p3+i) ⇡ e = ⇡ Re (1 p3 i)e + Re 24 2 12 ◆ ⇡e 2p3 ⇡ = (cos 2 p3 sin 2) + cos 4. 24 24 ⇤ Improper Integrals Involving Logarithms 1 ln x Example (6.18). Evaluate dx where a > 0. x2 + a2 Z0 Solution. This integral exists in the ordinary sense since both integrals on the right side of the following converge: 1 1 ln x ln x 1 ln x dx = dx + dx. x2 + a2 x2 + a2 x2 + a2 Z0 Z0 Z1 Because 0 1 ln x ln( x) 2 2 dx = 2 2 dx 0 x + a x + a Z Z 1 log ⇡/2 z we consider dz where C is the contour below. z2 + a2 IC 168 6. RESIDUE THEORY The integrand has simple poles at ai, so that ± log ⇡/2 z log ⇡/2 z (z ai) log ⇡/2 z dz = 2⇡i Res , ai = 2⇡i lim z2 + a2 z2 + a2 z ai (z ai)(z + ai) IC ! ⇡ ⇡ ⇡i = log ⇡/2(ai) = ln a + . a a 2 On the semicircle , set z = Re✓i, 0 ✓ ⇡ = ) log ⇡/2 z ln R + ✓i ln R + ⇡ | | provided R > a and R > 1 = z2 + a2 R2 a2 R2 a2 ) log ⇡/2 z ln R + ⇡ dz (⇡R) 0 as R . z2 + a2 R2 a2 ! ! 1 I ✓i On the semicircle 0, set z = re , ⇡ ✓ 0 = ) log ⇡/2 z ln r + ✓i ln r + ⇡ | | | | = z2 + a2 a2 r2 a2 r2 ) log ⇡/2 z ln r + ⇡ dz | | (⇡ r) 0 as r 0. z2 + a2 a2 r2 ! ! I 0 Referring to the equation at the top of this page, ⇡ ⇡i log ⇡/2 z 1 log ⇡/2 x ln a + = lim dz = dx a 2 r 0 z2 + a2 x2 + a2 R! C !1 I Z 1 0 ln( x) + ⇡i 1 ln x = 2 2 dx + 2 2 dx = x + a 0 x + a 1 Z 0Z 1 ln x ⇡i = 2 2 2 dx + 2 2 dx. 0 x + a x + a Z Z 1 Then real parts give 1 ln x ⇡ dx = ln a. x2 + a2 2a Z0 ⇤ 6.3. MORE IMPROPER INTEGRALS BY RESIDUES 169 Improper Integrals Involving Fractional Powers 1 1 Example (6.19). Evaluate dx, where 0 < a < 1. xa(x + 1) Z0 Solution. To evaluate this improper integral by contour integration, we 1 replace the real integrand with the conplex za(z + 1) . Because za, for fractional a, is multiple-valued, we must choose some branch of the function. This is dictated by the choice of contour. To obtain⇥ the required⇤ real improper integral, part of the contour should include a part of the positive real axis. To avoid the branch point, we include a small circle (of radius r) around z = 0. We also need a circle of large radius R centered at z = 0. This gives us the curves on the left below, but they do not form a closed contour. Our final choice of contour C is that shown on the right. It consists of a portion of the circle of radius r < 1 centered at the origin, and two horizontal line segmennts joining this circle to a portion of the large circle with radius R > 1.We choose the branch za = ea log0 z with branch cut along the positive real axis. To get the required real integral, we eventually take limits as r 0 and R . Since the integrand has a simple pole at z = 1 interior to !C, ! 1 170 6. RESIDUE THEORY 1 1 dz = 2⇡i Res , 1 za(z + 1) za(z + 1) IC 2⇡i 2⇡i 2⇡i a⇡i = = = = 2⇡ie . ( 1)a ea log0( 1) ea⇡i Breaking C into its four component parts, a⇡i 1 1 2⇡ie = dz + dz za(z + 1) za(z + 1) Z 1 Z 2 1 1 + dz + dz. za(z + 1) za(z + 1) Z 3 Z 4 On 1, 1 1 1 1 = = = za(z + 1) (z + 1)ea log0 z (z + 1)ea(ln R+arg0 z i) Ra(R 1) ) 1 1 dz (2⇡R) 0 as R . za(z + 1) Ra(R 1) ! ! 1 Z 1 On 2, 1 1 1 1 = = = za(z + 1) (z + 1)ea log0 z (z + 1)ea(ln r+arg0 z i) ra(r 1) ) 1 1 dz (2⇡r) 0 as r . za(z + 1) ra(r 1) ! ! 1 Z 1 When limits are taken on the result of Cauchy’s residue theorem as r 0 and R , ! ! 1 a⇡i 1 1 2⇡ie = lim dz + lim dz r 0 za(z + 1) r 0 za(z + 1) R! 3 R! 4 !1 Z !1 Z 1 1 = lim dz + lim dz. r 0 (z + 1)ea log0 z r 0 (z + 1)ea log0 z R! 3 R! 4 !1 Z !1 Z 6.3. MORE IMPROPER INTEGRALS BY RESIDUES 171 Each of these last two limits are along the positive real axis. Since the integral along 3 is in the first quadrant, the limit of the argument in log0 z is 0. Since 4 is in the fourth quadrant, the limit of the argument in log0 z is 2⇡. Thus, when limits are taken, 0 a⇡i 1 1 1 2⇡ie = a ln x dx + a ln x+2⇡i dx 0 (x + 1)e (x + 1)e Z Z1 1 1 1 1 = dx dx xa(x + 1) xa(x + 1)e2a⇡i Z0 Z0 2a⇡i 1 1 = (1 e ) dx. xa(x + 1) Z0 Solving this equation for the required integral, a⇡i 1 1 2⇡ie 2⇡i 2⇡i ⇡ dx = = = = . xa(x + 1) 1 e 2a⇡i ea⇡i e a⇡i 2i sin a⇡ sin a⇡ ⇤ Z0