Topics to Review Laurent series and Laurent coefficients from Section 4.5 are used to define residues in Section 5.. auchy s theorem for multiply connected regions Theorem 6, Section 3.4 is the basis for most integrals in this chapter. Section 5.3 uses the estimates on the path integral from Theorem 2, Section 3.2. Section 5.6 is based on the residues of the cotangent, which are presented in Example 4, Section 5.. Looking Ahead Section 5. contains the main result of this chapter along with some basic applications that illustrate its use in computing integrals. Section 5.2 deals with the integrals of special trigonometric functions that can be evaluated directly from the residue theorem. Sections 5.3, 5.4, and 5.5 are more illustrative of the residue method. The integrals that we compute are classical and arise in many applications. The computations require the residue theorem and additional estimates, which vary in difficulty from simpler ones in Section 5.3 to more difficult ones in Section 5.5. Section 5.6 is an interesting application to series. Sections 5.5 and 5.6 can be skipped without affecting the continuity of the course. Section 5.7 contains many interesting theoretical properties, whose proofs use at some stage the residue theorem. While this section is optional, it is strongly recommended in a course that stresses mathematical proofs. Interesting applications are presented in the exercises of that section. 5 RESIDUE THEORY After having thought on this subject, and brought together the diverse results mentioned above, I had the hope of establishing on a direct and rigorous analysis the passage from the real to the imaginary; and my research has lead me to this Memoire. -Augustin Louis auchy [Writing about his Memoire of 84, which contained the residue theorem and several computations of real integrals by complex methods.] In the previous four chapters, we introduced complex numbers and complex functions and studied properties of the three essential tools of calculus: the derivative, the integral, and series of complex functions. In this and the next chapter, we use these results to derive some exciting applications of complex analysis. The applications of this chapter are based on one formula, known as auchy s residue theorem. We have already used this result many times before when computing integrals. Here we will highlight the main ideas behind it and devise new techniques for computing important integrals that arise from Fourier series and integrals, Laplace transforms, and other applications. For example, integrals like sin x x dx, 0 cos xe x2 dx are very difficult to compute using real variable techniques. With the residue theorem and some additional estimates with complex functions, the computations of these integrals are reduced to simple tasks. In Section 5.7 an optional section, we will use residues to expand our knowledge of analytic functions. We will use integrals to count the number of eros of analytic functions and to give a formula for the inverse of an analytic functions. Theoretical results, such as the open mapping property, will be derived and will be used to obtain a fresh and different perspective on concrete results such as the maximum modulus principle. The residue theorem was discovered around 84 stated explicitely in 83 by auchy as he attempted to generalie and put under one umbrella the computations of certain special integrals, some of them involving complex substitutions, that were done by Euler, Laplace, Legendre, and other mathematicians.
Section 5. auchy s Residue Theorem 287 5. auchy s Residue Theorem Suppose that f has an isolated singularity at 0. We know from Theorem, Section 4.5, that f has a Laurent series in an annulus around 0 : for 0 < 0 <R, f = + a 2 0 2 + a 0 + a 0 + a 0 +a 2 0 2 +. Furthermore, the series can be integrated term by term over any path that lies in the annulus 0 < 0 <R. Let r 0 be any positively oriented circle that lies in 0 < 0 <R. If we integrate the Laurent series term by term over r 0 and use the fact that r 0 0 n d = 0 if n and r 0 0 d =2πi, we find r 0 f d = a 2πi; hence a = f d. 2πi r 0 THEOREM AUHY S RESIDUE THEOREM The coefficient a is called the residue of f at 0 and is denoted by Res f, 0 or simply Res 0 when there is no risk of confusing the function f. With the concept of residue in hand, we can state our main result. Let be a simple closed positively oriented path. Suppose that f is analytic inside and on, except at finitely many isolated singularities, 2,..., n inside. Then n 2 f d =2πi Res f, j. j= Proof Take small circles rj j j =, 2,..., n that do not intersect each other and are contained in the interior of Figure. Apply auchy s integral theorem for multiple simple paths Theorem XX, Section 3.4 and get f d = n j= rj f d =2πi n Res j, j= where the last equality follows from. So 2 holds. Figure If is negatively oriented, we need to add a negative sign to the right side of 2. Theorerm reduces the evaluation of certain integrals to computating residues. The computation of the residue will depend on the type of singularity of the functioon f, as we now illustrate with useful results.
288 hapter 5 Residue Theory PROPOSITION RESIDUE AT A SIMPLE POLE i Suppose that 0 is an isolated singularity of f. Then f has a simple pole at 0 if and only if 3 Res f, 0 = lim 0 0 f 0. ii If f = p q, where p and q are analytic at 0, p 0 0, and q has a simple ero at 0, then 4 Res p q, p 0 0 = q 0. Proof i By Theorem 7, Section 4.6, 0 is a pole of order if and only if the Laurent series of f at 0 is f = a + a 0 + a 0 +a 2 0 2 + = a + h, 0 0 where a 0 and h is the analytic power series part of the Laurent series. Then, 0 f =a + 0 h, and i follows upon taking the limit as 0. To prove ii, note that f has a simple pole at 0. Using i and q 0 = 0, we have Res p q, 0 = lim 0 p 0 q = lim 0 p lim 0 0 q q 0 = p 0 q 0. EXAMPLE An application of the residue theorem Let be a simple closed positively oriented path such that, i, and i are in the interior of and is in the exterior of Figure 2. Find d 4. Figure 2 The path and the poles of f in Example. Solution The function f = 4 has isolated singularities at = ± and ±i. Three of these are inside, and according to 2 we have d 5 4 =2πi Res + Res i + Res i. We have 4 = + i + i, so ± and ±i are simple roots of the polynomial 4 = 0. Hence f = 4 has simple poles at ± and ±i Theorem 7v, Section 4.6. Let 0 denote any one of the points, ±i. Because 0 is a simple pole, we have from 3, 6 lim 0 0 f =a = Res 0. Using the factoriation 4 = + i + i, we have at 0 = Res = lim = 4 = lim + i + i = = 4. + i + i
Section 5. auchy s Residue Theorem 289 Similarly, at 0 = i, we have Res i = lim i i 4 = + + i = i =i 4, and at = i, Res i = lim + i i 4 = + i = i = i 4. Plugging these values into 5, we obtain d 4 = πi 2. EXAMPLE 2 Residue at a removable singularity Let be a simple closed positively oriented path such that is in the interior of and is in the exterior of Figure 3. Find sinπ I = 2 d. Figure 3 FIX THIS FIGURE FOR Example 2. THEOREM 2 RESIDUE AT A POLE OF ORDER m Solution The function f = sinπ 2 has isolated singularities at = ±; only is inside. Since lim sinπ 2 = lim sinπ + =0, it follows from Theorem 6, Section 4.6, that is a removable singularity of f. Thus, the Laurent series of f at 0 = has no negative powers of, and, in particular, a = 0. Hence Res f, = a = 0, and so I = 0. You should also verify that I = 0 by using auchy s integral formula. Example 2 brings up the following simple observation: If 0 is a removable singularity, then Res f 0 = 0. For poles of higher order the situation is more complicated. Suppose that 0 is a pole of order m of f. Then the residue of f at 0 is 7 Res f, 0 = lim 0 m! d m d m [ 0 m f], where as usual the derivative of order 0 of a function is the function itself. You should check that for m = formula 7 reduces to 3. Proof By the Laurent series characteriation of poles Theorem 7, Section 4.6, f = a m 0 m + + a 0 + a 0 + a 0 +a 2 0 2 +. Multiply by 0 m, then differentiate m times to obtain d m d m [ 0 m f] = m!a + m!a 0 0 + m +! a 0 2 +. 2
290 hapter 5 Residue Theory Take the limit as 0, and get lim 0 which is equivalent to 7. d m d m [ 0 m f] = m!a +0, EXAMPLE 3 omputing residues Let be the simple closed path shown in Figure 3. a ompute the residues of f = 2 2 +π 2 2 sin at all the isolated singularities inside. b Evaluate 2 2 + π 2 2 sin d. Figure 3 The path and the poles of f in Example 2. Solution a Three steps are involved in answering this question. Step : Determine the singularities of f inside. The function f = 2 2 +π 2 2 sin is analytic except where 2 + π 2 = 0 or sin = 0. Thus f has isolated singularities at ±iπ and at kπ where k is an integer. Only 0 and iπ are inside. Step 2: Determine the type of the singularities of f inside. Let us start with sin the singularity at 0. Using lim 0 =, it follows that lim 0 =, and so sin lim f = lim 0 0 sin = 0 = 0. 2 + π 2 2 By Theorem 5, Section 4.6, f has a removable singularity at 0 = 0. To treat f = +iπ2 iπ 2 sin 2 f the singularities at iπ, we consider. learly has a ero of order 2 at iπ, and so by Theorem 7v, Section 4.6, f has a pole of order 2 at iπ. Step 3: Determine the residues of f inside. At 0, f has a removable singularity, so a = 0, and hence the residue of f at 0 is 0. At iπ, we apply Theorem 2, with m = 2, 0 = iπ. Then Res iπ = lim iπ = lim iπ d [ iπ 2 f ] d [ d 2 ] d + iπ 2 sin 2 + iπ sin 2 + iπ cos + 2 sin = lim iπ + iπ 3 sin 2 2 sinh π + πcosh π sinh π = 4π sinh 2 = π 4π sinh π + cosh π 4π sinh 2 π, where the last line follows by plugging = iπ into the previous line and using siniπ =i sinh π and cosiπ = cosh π. b Using Theorem and a, we obtain 2 2 + π 2 2 sin d = 2πi Res 0 + Res iπ = i 2 sinh π + cosh π 2 sinh 2. π
Section 5. auchy s Residue Theorem 29 EXAMPLE 4 Residues of the cotangent a Let k be an integer. Show that Res cotπ,k = π. b Suppose that f is analytic at an integer k. Show that 8 Res f cotπ,k = π fk. Figure 4 The path and the poles of f cotπ in Example 4c. PROPOSITION 2 RESIDUE OF AN EVEN FUNTION c Evaluate cotπ + 4 d, where is the positively oriented rectangular path shown in Figure 4. Solution a We know that the eros of φ = sinπ are precisely the integers. Also, since φ k =π coskπ 0, it follows from Theorem, Section 4.6, that all the eros of sinπ are simple eros. Hence cotπ = cosπ sinπ has simple poles at the integers. To find the residue at k, we use Proposition ii. We have Res cotπ, k = Res cosπ sinπ,k = coskπ d d sinπ = =k π. b This is immediate from a and Proposition ii: Take p = f cosπ and q = sinπ. c Since + 4 is nonero inside and cotπ has simple poles at the integers, it follows that cotπ + has two simple poles inside at = 0 and =. Applying 4 Theorem and using 8 with f = + to compute the residues, we find 4 cotπ d = 2πi Res cotπ, 0 + Res cotπ +4 +4 + 4, = 2πi π + 0 + =2i + =3i. 4 π + 4 2 So far the examples that we treated involved residues at poles of finite order. There is no formula like 7 for computing the residue at an essential singularity. We have to rely on various tricks to evaluate the coefficient a in the Laurent series expansion. We illustrate with several examples, starting with a useful observation. Suppose that 0 is an isolated singularity of an even function f. Res f, 0 = 0. Then Proof We have to show that the a the Laurent series coefficient of f at 0 is 0. Write f = n= a n n, where 0 < <r. Substitue for and use the fact that f is even so f =f. Then n= a n n = n= n a n n, and by the uniqueness of the Laurent series, it follows that n a n = a n, which implies that a n = 0 if n is odd; in particular, a = 0. EXAMPLE 5 Residue at 0 of an even function ompute Res e 2 cos, 0.
292 hapter 5 Residue Theory Solution The function e 2 cos is even and has an isolated essential singularity at 0. By Proposition 2, Res e 2 cos, 0 = 0. Multiplication of series is often useful in computing residues at a singularity, including essential singularities. EXAMPLE 6 Multiplying series by a polynomial ompute the residues of 2 sin at = 0. Solution From the Laurent series we get and so Res 2 sin, 0 = 3!. sin = 3! 2 sin = 3! 3 + 5! + 5! 5, 3, Figure 5 The path R for Exercises 5 and 20. EXAMPLE 7 Using auchy products e Find the residue of at = 0. 2 + Solution The given function has an essential singularity at = 0. To compute the coefficient a in its Laurent series around 0, we use two familiar Taylor and Laurent series as follows. We have, for 0 < <, e 2 + = 2 + e = 2 + 4 + + 2! + 2 3! + 3 By properties of Taylor and Laurent series, both series are absolutely convergent in 0 < <. So we can multiply them term by term using a auchy product. ollecting all the terms in, we find that e Res 2 +, 0 = a = 3! + = sin. 5! Exercises 5. In Exercises 2, find the residue of the given function at all its isolated singularities.. + + 2. 3. +e 2 +2 +2 2 + 2 sin 2 3 4. 2 2 5. 6. cos + +3i 3 7. 8. cotπ 9. cscπ + sin + 0. sin. e + 2. cos sin In Exercises 3 26, evaluate the given path integral. The path R in Exercises 5 and 20 is shown in Figure 5..
3. 5. 7. 9. 2. 23. 25. 26. 0 R Section 5. auchy s Residue Theorem 293 2 +3 2 3 d. 4. 0 5 d + i e d i2 i 3 6. i 30 2 + 3 3i 2 6i d 32 0 40 0 0 0 /2 0 d 2 0 8. tan d 20. e 2 d 6 22. 4 e + 2 d 24. sin 6 d 4 e d 30 R 2 + 2 d d +e π 0 3/2 0 cos e 2 d 2 cotπ d 27. a Prove that if f has a simple pole at 0 and g is analytic at 0, then Res fg, 0 =g 0 Res f, 0. b Use a to prove 8. 28. Show that Res f+g, 0 = Res f, 0 + Res g, 0. 29. Residues of the cosecant. a Show that cscπ has simple poles at the integers. b For an integer k show that c Res cscπ, k = k π. Suppose that f is analytic at an integer k. Show that 30. Use Exercise 29 to compute a cscπ d, b 25/2 0 Res f cscπ, k = k π fk. 25/2 0 cscπ + 2 d. 3. Explain how the residue theorem implies auchy s theorem Theorem 5, Section 3.4 and auchy s integral formula Theorem 2, Section 3.6. 32. Suppose that f has an isolated singularity at 0. Show that Res f, 0 = 0. 33. onsider the Laurent series expansions in an annulus around 0, f = n= a n 0 n and g = n= b n 0 n.