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Residue Thry 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− .
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