17 Residue Theory
“Residue theory” is basically a theory for computing integrals by looking at certain terms in the Laurent series of the integrated functions about appropriate points on the complex plane. We will develop the basic theorem by applying the Cauchy integral theorem and the Cauchy integral formulas along with Laurent series expansions of functions about the singular points. We will then apply it to compute many, many integrals that cannot be easily evaluated otherwise. Most of these integrals will be over subintervals of the real line.
17.1 Basic Residue Theory The Residue Theorem Suppose f is a function that, except for isolated singularities, is single-valued and analytic on some simply-connected region R . Our initial interest is in evaluating the integral
f (z) dz . C I 0 where C0 is a circle centered at a point z0 at which f may have a pole or essential singularity. We will assume the radius of C0 is small enough that no other singularity of f is on or enclosed by this circle. As usual, we also assume C0 is oriented counterclockwise. In the region right around z0 , we can express f (z) as a Laurent series
∞ k f (z) ak (z z0) , = − k =−∞X and, as noted earlier somewhere, this series converges uniformly in a region containing C0 . So
∞ k ∞ k f (z) dz ak (z z0) dz ak (z z0) dz . C = C − = C − 0 0 k k 0 I I =−∞X =−∞X I But we’ve seen k (z z0) dz for k 0, 1, 2, 3,... C − = ± ± ± I 0
3/10/2014 Chapter & Page: 17–2 Residue Theory
before. You can compute it using the Cauchy integral theorem, the Cauchy integral formulas, or even (as you did way back in exercise 14.14 on page 14–17) by direct computation after parameterizing C0 . However you do it, you get, for any integer k ,
k 0 if k 1 (z z0) dz 6= − . C0 − = ( i2π if k 1 I =− So, the above integral of f reduces to
∞ 0 if k 1 f (z) dz ak 6= − i2πa 1 . C = = − 0 k (i2π if k 1) I =−∞X =− This shows that the a 1 coefficient in the Laurent series of a function f about a point z0 completely determines the− value of the integral of f over a sufficiently small circle centered at z0 . This quantity, a 1 , is called the residue of f at z0 , and is denoted by −
a 1,z0 or Resz0 ( f ) or − ···
In fact, it seems that every author has their own notation. We will use Resz0 ( f ) . If, instead of integrating around the small circle C0 , we were computing
f (z) dz C I where C is any simple, counterclockwise oriented loop in R that touched no point of singularity of f but did enclose points z0 , z1 , z2 , … at which f could have singularities, then, a consequence of Cauchy’s integral theorem (namely, theorem 15.5 on page 15–7) tells us that
f (z) dz f (z) dz C = C k k I X I where each Ck is a counterclockwise oriented circle centered at zk small enough that no other point of singularity for f is on or enclosed by this circle. Combined with the calculations done just above, we get the following:
Theorem 17.1 (Residue Theorem) Let f be a single-valued function on a region R , and let C be a simple loop oriented counter- clockwise. Assume C encircles a finite set of points z0, z1, z2,... at which f might not be { } analytic. Assume, further, that f is analytic at every other point on or enclosed by C . Then
f (z) dz i2π Reszk ( f ) . (17.1) C = k I X In practice, most people just write equation (17.1) as
f (z) dz i2π sum of the enclosed residues . C = × I The residue theorem can be viewed as a generalization of the Cauchy integral theorem and the Cauchy integral formulas. In fact, many of the applications you see of the residue theorem can be done nearly as easily using theorem 15.5 (the consequence of the Cauchy integral theorem used above) along with the Cauchy integral formulas. Basic Residue Theory Chapter & Page: 17–3
Computing Residues
Remember, what we are now calling the residue of a function f at z0 is simply the value of a 1 in the Laurent series expansion of f , −
∞ k f (z) ak (z z0) , = − k =−∞X right around z0 ,and,forthis a 1 tobenonzero, f must haveeither apoleor essential singularity − at z0 . Sothediscussionaboutsuchsingularitiesinsection16.3 applies both for identifying where a function may have residues and for computing the residues. The basic approach to computing the residue at z0 is be to simply find the above Laurent series. Then
Resz0 ( f ) a 1 . = − This may be necessary if f has an essential singularity at z0 . If f has a pole of finite order, say, of order M , then we can use formula (16.8) on page 16–14 for a 1 , − M 1 1 d − M Resz0 ( f ) a 1 (z z0) f (z) . (17.2) = − = (M 1) dzM 1 − − ! − z z0 =
If the pole is simple (i.e., M 1), this simplifies to =
Resz0 ( f ) a 1 (z z0) f (z) z z0 . (17.3) = − = − | = Often, we may notice that g(z) f (z) = z z0 − for some function g which is analytic and nonzero at z0 . In this case, we clearly have a simple pole, and formula (17.3), above, clearly reduces to
Resz ( f ) g(z0) . 0 = This will make computing residues very easy in many cases. More generally, from our earlier discussion of poles, we know that if
g(z) f (z) M (17.4) = (z z0) − for some function g which is analytic and nonzero at z0 , then f has a pole of order M at z0 . In this case, formula (17.2) reduces to
1 (M 1) Resz ( f ) g − (z0) . (17.5) 0 = (M 1) − ! Keep in mind that, if g(z0) 0, then the pole of = g(z) f (z) M = (z z0) − has order less than M , and a little more work will be needed to determine the precise order of the pole and the corresponding residue. version: 3/10/2014 Chapter & Page: 17–4 Residue Theory
17.2 “Simple” Applications The main application of the residue theorem is to compute integrals we could not compute (or don’t want to compute) using more elementary means. We will consider some of the common cases involving single-valued functions not having poles on the curves of integration. Later, we will add poles and deal with multi-valued functions.
2π Integrals of the form f (sin θ, cos θ) dθ Z0 Suppose we have an integral over (0, 2π) of some formula involving sin(θ) and cos(θ) , say, 2π dθ . 2 cos θ Z0 + We can convert this to an integral about the unit circle by using the substitution z eiθ . = Note that z does go around the unit circle in the counterclockwise direction as θ goes from 0 to 2π . Under this substitution, we have dz d eiθ ieiθ dθ izdθ . = = = So, 1 1 dθ dz iz− dz . = iz = − For the sines and cosines, we have iθ iθ 1 1 e e z /z z z cos θ + − + + − = 2 = 2 = 2 and iθ iθ 1 1 e e z /z z z sin θ − − − − − . = 2i = 2i = 2i These substitutions convert the original integral to an integral of some function of z over the unit circle, which can then be evaluated by finding the enclosed residues and applying the residue theorem.
!◮Example 17.1: Let’s evaluate 2π dθ . 2 cos θ Z0 + Letting C denote the unit circle and applying the substitution z eiθ , as described above, = we get 2π dθ iz 1 dz − − 1 1 0 2 cos θ = C 2 z z Z + I + 2 + − 1 2z iz− −1 dz = C 2z · 2 z z 1 I + 2 + − i2 1 − dz 2i dz . 2 2 = C 4z z 1 = − C z 4z 1 I + + I + + “Simple” Applications Chapter & Page: 17–5
Letting 1 f (z) = z2 4z 1 + + and applying the residue theorem, the above becomes
2π dθ 2i i2π sum of the residues of f (z) in the unit circle 2 cos θ = − × × Z0 + 4π sum of the residues of f (z) in the unit circle . = × To find the necessary residues we must find where the denominator of f (z) vanishes,
z2 4z 1 0 . + + = Using the quadratic formula, these points are found to be
4 √42 4 z − ± − 2 √3 . ± = 2 = − ± So 1 f (z) . = z 2 √3 z 2 √3 − − + − − − Now, since √3 1.7 , ≈ z 2 √3 2 1.7 0.3 | +| = − + ≈ |− + | =
while z 2 √3 2 1.7 3.7 . | −| = − − ≈ |− − | =
Clearly, 2 √3 is the only singular point of f (z) enclosed by the unit circle, and the − + singularity there is a simple pole. We can rewrite f (z) as
g(z) 1 f (z) where g(z) . = z 2 √3 = z 2 √3 − − + − − − Thus,
1 1 Resz f g(z ) g( 2 √3) . + [ ] = + = − + = 2 √3 2 √3 = 2√3 − + − − − Plugging this back into the last formula obtained for our int egral, we get
2π dθ 4π sum of the residues of f (z) in the unit circle 2 cos θ = × Z0 + 4π Resz f = + [ ] 1 4π = 2√3 2π . = √3
version: 3/10/2014 Chapter & Page: 17–6 Residue Theory
Y Y
ΓR+ C R+
R IR R X IR − R R X C − R− ΓR−
(a) (b)
Figure 17.1: Closed curves for integrating on the X–axis when (a) ΓR Γ + IR C+ = R = + R and when (b) ΓR Γ − IR C− . = R =− + R
∞ Integrals of the form f (x) dx Z−∞ Now let’s consider evaluating ∞ f (x) dx , Z−∞ assuming that:
1. Except for a finite number of poles and/or essential singularities, f is a single-valued analytic function on the entire complex plane.
2. None of these singularities are on the X–axis.
3. One of the following holds: (a) zf (z) 0 as z . → | | → ∞ (b) f (z) g(z)eiαz where α > 0 and = g(z) 0 as z . → | | →∞ iαz (c) f (z) g(z)e− where α > 0 and = g(z) 0 as z . → | | →∞ Under these assumptions,we can evaluate the integral by constructing,for each R > 0,asuitable closed loop ΓR containing the interval ( R, R) . The integral over ΓR is computed via the − residue theorem, and R is allowed to go to . The conditions listed under the third assumption ∞ above ensure that the integral over that part of ΓR which is not the interval ( R, R) vanishes − as R , leaving us with the integral over ( , ) . →∞ −∞ ∞ The exact choice for the closed loop ΓR depends on which of the three conditions under assumption 3 is known to hold. If either (3a) or (3b) holds, we take
ΓR Γ + IR C+ = R = + R where IR is the subinterval ( R, R) of the X–axis and C+ is the semicirle in the upper − R half plane of radius R and centered at 0 (see figure 17.1a). In this case, with ΓR+ oriented “Simple” Applications Chapter & Page: 17–7
counterclockwise, the direction of travel on the interval IR is from x R to x R (the = − = normal ‘positive’ direction of travel on the X–axis) and, so,
R f (z) dz f (x) dx . IR = R Z Z− Since we are letting R and there are only finitely many singularities, we can always → ∞ assume that we’ve taken R large enough for ΓR+ to enclose all the singularities of f in the upper half plane (UHP). Then
i2π sum of the residues of f in the UHP f (z) dz = Γ Z R+ f (z) dz f (z) dz . = I + C Z R Z +R Thus,
R f (x) dx f (z) dz R = IR Z− Z i2π sum of the residues of f in the UHP f (z) dz . (17.6) = − C Z +R C To deal with the integral over R+ , first note that this curve is parameterized by
z z(θ) Reiθ where 0 θ π . = = ≤ ≤ So “ dz d Reiθ iReiθ dθ ” and = [ ]= π f (z) dz f Reiθ iReiθ dθ . C = Z +R Z0 Now assume assumption (3a) holds, and let
M(R) the maximum of zf (z) when z R . = | | | | = By definition, then, Reiθ f Reiθ M(R) ≤ Moreover, it is easily verified that assumption (3a) (that zf (z) 0 as z ) implies that → | | →∞ M(R) 0 as R . → →∞ So π lim f (z) dz lim f Reiθ iReiθ dθ R R C+ = 0 →∞ Z R →∞ Z π lim f Reiθ iReiθ dθ ≤ R →∞ Z0 π lim M(R) dθ lim M(R) π 0 . ≤ R = R = →∞ Z0 →∞ version: 3/10/2014 Chapter & Page: 17–8 Residue Theory
Thus, lim f (z) dz 0. (17.7) R C = →∞ Z +R Under assumption (3b),
f (z) dz g(z)eiαz dz C = C Z +R Z +R for some α > 0 . Now,
iαz iα(x iy) iαx αy αy e e + e e− e− , = = = which goes to zero very quickly as y . So it is certainly reasonable to expect equation → ∞ (17.7) to hold under assumption (3b). And it does — but a rigorous verification requires a little more space than is appropriate here. Anyone interested can find the details in the proof of lemma 17.2 on page 17–14. So, after letting R , formula (17.6) for the integral of f (x) on ( R, R) becomes →∞ − ∞ f (x) dx i2π sum of the residues of f in the UHP . (17.8) = Z−∞ On the other hand, if it is assumption (3c) that holds, then
iαz f (z) dz g(z)e− dz C = C Z +R Z +R for some α > 0. In this case, though,
iαz iα(x iy) iαx αy αy e− e− + e− e e , = = = which rapidly blows up as y gets large. So it is not reasonable to expect equation (17.7) to hold here. Instead, take ΓR Γ − IR C− = R = − + R C where R− is the semicirle in the lower half plane of radius R and centered at 0 (see figure 17.1b). (Observe that, this time, the direction of travel on IR is opposite to the normal ‘positive’ direction of travel on the X–axis.) Again, since we are letting R and there are only → ∞ finitely many singularities, we can assume that we’ve taken R large enough for ΓR− to enclose all the singularities of f in the lower half plane (LHP). Then
i2π sum of the residues of f in the LHP f (z) dz = Γ Z R− f (z) dz f (z) dz = IR + C− Z− Z R f (z) dz f (z) dz . = − I + C Z R Z −R Thus, R f (x) dx f (z) dz R = − IR Z− Z− i2π sum of the residues of f in the UHP f (z) dz = − + C Z +R “Simple” Applications Chapter & Page: 17–9
But, as before, it can be shown that f (z) 0 fast enough as z (on the lower half | | → | | → ∞ plane) to ensure that lim f (z) dz 0 . R C = →∞ Z −R So, after letting R , the above formula for the integral of f (x) on ( R, R) becomes →∞ − ∞ f (x) dx i2π sum of the residues of f in the LHP . (17.9) = − Z−∞ !◮Example 17.2: Let us evaluate the “Fourier integral”
ei2πx ∞ dx . 1 x2 Z−∞ + Here we have 1 f (z) g(z)ei2πz with g(z) . = = 1 z2 + Clearly, g(z) 0 as z and 2π > 0 ; so condition (3b) on page 17–6 holds. Thus, | | → | | →∞ we will be applying equation (17.8), which requires the residues of f from the upper half plane.1 By inspection, we see that
ei2πz ei2πz f (z) , = 1 z2 = (z i)(z i) + + − which tells us that the only singularities of f (z) are at z i and z i . Only i , though, = =− is in the upper half plane, so we are only interested in the residue at i . Since we can write f (z) as h(z) ei2πz f (z) with h(z) = z i = z i − + (and h(i) 0 ), we know f (z) has a simple pole at i and 6= i2πi 2π e e− Resi f h(i) . [ ] = = i i = 2i + Thus, applying equation (17.8)
ei2πx ∞ dx i2π sum of the residues of f in the UHP 1 x2 = Z−∞ + i2π Resi f = [ ] 2π e− 2π i2π πe− . = 2i =
1 Rather than memorize that “condition (3b) implies that (17.8) is used”,keep in mind the derivation and the fact that you want the integral over one of the semicircles to vanish as R . Write out the exponentialin terms of x and y and see if this exponentialis vanishing as y or as y →∞ . Then “rederive” the residue-based formula for the integral of interest using the semicircle→ in the+∞ upper half→ plane −∞ if the exponential vanishes as y and the semicircle in the lower half plane if the exponential vanishes as y . For our example → +∞ → −∞ ei2πz ei2π(x iy) ei2π e 2πy 0 as y . = + = − → → +∞ So we are using the upper half plane. version: 3/10/2014 Chapter & Page: 17–10 Residue Theory
Standard “Simple” Tricks Using Real and Imaginary Parts It is often helpful to observe that one integral of interest may be the real or imaginary part of another integral that may, possibly, be easier to evaluate. Do remember that
b b b b Re f (x) dx Re f (x) dx and Im f (x) dx Im f (x) dx . [ ] = [ ] = Za Za Za Za (If this isn’t obvious, spend a minute to (re)derive it.) In this regard, it is especially useful to observe that cos(X) Re ei X and sin(X) Im ei X . = = !◮Example 17.3: Consider cos(2πx) ∞ dx 1 x2 Z−∞ + Using the above observations and our answer from the previous exercise,
cos(2πx) ei2πx ∞ dx ∞ Re dx 1 x2 = 1 x2 Z−∞ + Z−∞ + i2πx ∞ e 2π 2π Re dx Re πe− πe− . = 1 x2 = = Z−∞ +
Clever Choice of Curve and Function The main “trick” to applying residue theory in computing integrals of real interest (i.e., integrals that actually do arise in applications) as well as other integrals you may encounter (e.g., other integrals in assigned homework and tests) is to make clever choices for the functions and the curves so that you really can extract the value of desired integral from the integral over the closed curve used. Some suggestions, such as were given on page 17–6 for computing certain integrals on ( , ) , can be given. In general, though, choosing the right curves and functions is a −∞ ∞ cross between an art and a skill that one just has to develop. A good example of using both clever choices of functions and curves, and in using real and imaginary parts, is given in computing the Fresnel integrals (which arise in optics).
!◮Example 17.4: Consider the Fresnel integrals
∞ cos x2 dx and ∞ sin x2 dx . Z0 Z0 Rather than use cos x2 and sin x2 directly, it is clever to use eix2 and the fact that