8 April 14, 2016 In this assignment, we will learn some techniques for calculating line integrals and as an application compute some real integrals with complex techniques. Simply connected domains. Recall that a domain is a connected open subset of C. We say that a domain U is simply connected if U has no holes. For example, D = {z : |z| < 1} (the unit disk) is simply connected, while D = {z : 0 < |z| < 1} (the punctured unit disk) is not simply connected as it has no holes. This concept is very intuitive, so we will not worry to much about dening it rigorously, but if you are curious, here is one possible denition: Denition 1. A domain U C is simply connected if for every point p which is not in U there is a continuous path starting at p and tending to which always lies outside of U . More formaally: there is a continuous path : [0, ) C \\ U such that (0) = p and limt (t) = Coming back to the punctured disk D dened above, it is not simply connected, since any continuous path from 0 to must pass through D . Exercise 1. Which of the following domains are simply connected (you do not need to give a proof)? (a) C \\ (, 0]. (b) C \\ [0, 1]. (c) {z C : 1 < |z 1| < 2}. (d) {z C : Im(z) > 0 and |z i| > 1/2}. (e) {z C : Im(z) > 0 and |z| < 1/2}. Simple connectivity is important because it often tells us when line integrals are 0. A path : [a, b] C is called closed or a loop if (a) = (b). In other words begins and ends at the same point (for example (t) = eit for t [0, ]. Theorem 1 (Cauchy's theorem). Let U C be a domain. The following are equivalent: 1 1. U is simply connected. 2. f (z) dt = 0 for any complex analytic function f on U and any loop in U . 3. For every complex analytic function f on U , there is a complex analytic function F on U such that F = f . The F in the last part of Cauchy's theorem is called an antiderivative of f . In single variable calculus, you learned that every continuous function on an interval has an antiderivative (this is part of the Fundamental Theorem of Calculus). For complex analytic functions, this is no longer true. Example 1. We have seen earlier that dz = 2i, z where is a circle of radius R centered at 0. If f (z) = 1/z had an antiderivative on U , this integral would have to be 0. Therefore 1/z has no antiderivative on U. On V = C \\ (, 0], Cauchy's theorem guarantees that 1/z has an antiderivative. If F (z) is a branch of log(z) on V , then F (z) = 1/z. The branches of the logarithm on V are all possible antiderivatives of 1/z. Exercise 2. Let U C be a simply connected domain and f (z) complex analytic on U . Let : [0, 1] U and : [0, 1] U be two curves which begin and end at the same point (that is, (0) = (0) and (1) = (1). Use Cauchy's theorem to show that f (z) dz = f (z) dz. This exercise can be rephrased as saying that on a simply connected domain, a line integral of a complex analytic function f (z) really only depends on the endpoints of . The residue theorem. Cauchy's theorem tells us that integrals of complex analytic functions over loops are always zero in a simply connected domain. On the other hand, we have seen that dz dz = 2i when is a circle centered z at zero (with the counterclockwise orientation). This does not violate Cauchy's theorem since the domain of 1/z is not simply connected (it has a hole at 0). The residue theorem is an easy way to evaluate integrals when a function is complex analytic on a simply connected domain, except for nitely many isolated singularities. Here is the setup of the residue theorem: Let D C be a simply connected domain whose boundary is a piecewisesmooth closed curve . We write D for D together with its boundary (this is called the closure of D. Some pictures of possible domains are pictured below: We give the counterclockwise orientation as indicated by the arrow (this is the usual orientation which we will always use unless said otherwise). Here's 2 Figure 1: Simply connected domains another way to think of this orientation: if you stand inside D and face towards the boundary, then the arrow is pointing to your left. Let f (z) be a complex analytic function on D (remember this means that f is complex dierentiable on every point of D), except for nitely many isolated singularities inside p1 , . . . , pn inside D. Recall that the residue of f (z) at pj is the coecient of the 1 term of the Laurent series of f (z) centered at the singularity pj . We write Respj f (z) for the residue of f (z) at pj . Theorem 2 (The Residue Theorem). In the above setup, n f (z) dz = 2i Respj f (z). j=1 Stated more briey: if encloses a simply connected region D, and f (z) is complex analytic on D and its boundary except for nitely many poles, then f (z) dz is 2i times the sum of the residues at the poles enclosed by . Note that it is important in the residue theorem that has the counterclockwise orientation. If you use the opposite orientation, then the sign of the integral is changed. z Example 2. Let f (z) = 1+z2 and is the circle of radius 2 centered at 0 with the counterclockwise orientation. There are two poles enclosed by at i and 3 i. We compute their residues: Resi f (z) = lim zi z(z i) z i 1 = lim = = . 2+1 zi z + i z 2i 2 A similar computation shows that Resi f (z) = 1 . Therefore, 2 z dz = 2i z2 + 1 1 1 + 2 2 = 2i 5 1 Example 3. Let be the square with vertices 1 i, 2 i, 5 + i, 2 + i, and 2 2 4 z let f (z) = sin(z) . Let's evaluate f (z) dz. The function f (z) has isolated singularities at every integer, since these are the zeros of sin(z). The curve encloses the singularities at 0, 1, 2, so we must compute the residues at these points. Since 0 is a zero of the numerator and a simple zero of sin(z) (since its derivative doesn't vanish there), 0 is a removable singularity. Recall at a removable singularity, the Laurent series has no negative terms, in particular Res0 f (z) = 0. Every other integer point is a simple pole of f (z) since the numerator doesn't vanish. We calculate these residues: z4 n4 n4 Resn = = (1)n . sin(z) cos(n) Here we used Rule 3 for computing residues from Homework 6. By the Residue Theorem: z4 1 16 dz = 2i 0 + sin(z) = 30i. Exercise 3. Compute the following line integrals: (a) 1 dz, 2 z+z where is the circle of radius 3 centered at 0. (b) ez dz, 2 (z 1) where is the circle of radius 3 centered at 0. (c) z dz, + 3z + 1 where is the circle of radius 1 centered at 0. z2 (d) z 2 (z 4z 5 dz, 1)(z 3) where is the circle of radius 1 centered at 1/2 4 Trigonometric integrals. Line integrals are useful even if you only care about calculus of real functions. Many real integrals can be expressed as line integrals and then evaluated easily using the Residue Theorem. Example 4. Let's use the residue theorem to evaluate the integral 2 0 d a + cos() for a > 1. (1) We will convert this to an integral on the unit circle with the parameterization (t) = eit . Recall by Euler's formula, ei + ei , and 2 ei ei . sin() = 2i cos() = We use this to write f () = z = ei : 1 a+cos() as a function of z by making the substitution 1 a + (ei + ei )/2 2 = 2a + z + z 1 2z = 2 z + 2az + 1 f () = (2) (3) (4) We write d in terms of dz by dierentiating both sides of z = ei : dz = iei d = izd dz d = i . z (5) (6) Substituting back into (1) gives: 2 0 2z dz dz z 2 + 2az + 1 iz 2 dz . = i z 2 + 2az + 1 d = a + cos() (7) (8) To check that this works, you can verify that these two integrals are the same by substituting back z = ei into the last integral, using the denition of a line integral. Now that we have turned our original problem into a line integral, we eval1 uate it using the residue theorem. The function g(z) = z2 +2az+1 has two poles at the zeros of z 2 + 2az + 1. Using the quadratic formula, we nd the zeros of this polynomial: z1 = a + a2 1 and z2 = a + 5 a2 1. These are the two poles of g(z). To apply the residue theorem, we need to nd which of these poles lie inside the unit circle. First, z1 = a a2 1 < a < 1, since a > 1 by assumption and a2 1 > 0 (the square root of any positive real number is positive). Since z1 is real and less than one, it lies outside the unit circle. Note that z1 z2 = 1, so z2 = z1 . Since z1 < 1, this tells us that 1 < z2 < 0, 1 so z2 lies in the unit circle. Now we know that g(z) has only one pole at inside the unit circle, at z2 . To compute the integral, it remains to compute the residue at this pole. Using Rule 3 from Homework 6 for computing the residue gives us: Resz2 1 1 = z 2 + 2az + 1 2z + 2a z=z2 1 = 2z2 + 2a 1 = . 2 a2 1 Now by the Residue theorem, 2 i dz 2 1 = 2i z 2 + 2az + 1 i 2 a2 1 2 . = a2 1 We conclude that for a > 1, 2 0 1 d = a + cos() 2 a2 1 Exercise 4. In this exercise, we will evaluate the integral 2 0 dz d. 5 + 4 sin() (a) Show that the above integral is equal to: f (z) dz = 2z 2 dz + 5iz 2 where is the unit circle parameterized by (t) = eit (see (4) and (6) above). (b) Find the poles of f (z) above. (c) Which poles lie inside the unit circle? (d) Find the residue at each pole inside the unit circle. (e) Use the residue Theorem to evaluate the above integral. 6 Integrals along the real axis. Many improper integrals along the real axis can be evaluated using the Residue Theorem. Example 5. Let's compute, dx , 1 + x2 by considering a line integral on the boundary of the upper half disk DR = {z : |z| < R and Im(z) > 0}, dz , (9) 1 + z2 DR (the notation DR means the curve bounding DR with the usual counterclockwise orientation). We can write the curve as DR = R + IR , where R is the upper-half circle parameterized by R (t) = Reit t [0, ], and IR is the real interval [R, R], parameterized by t [R, R]. IR (t) = t In those terms, the integral (9) becomes DR dz = 1 + z2 IR R = R dz + 1 + z2 R dx + 1 + x2 R dz 1 + z2 dz 1 + z2 (10) (11) (note the last equality is just the denition of the line integral over IR ). We use the Residue Theorem to compute the line integral (9). The function 1 1+z 2 has simple poles at i and i. Only i lies inside DR (for R > 1). We nd the residue at i: 1 1 1 Resz=i = = . 1 + z2 2z z=i 2i Therefore, by the Residue Theorem, DR dz = , 1 + z2 (12) as long as R > 1. Now taking the limit of both sides of (11) as R , and plugging in (12) to the integral on the left, we get, = dx + lim 1 + x2 R R dz . 1 + z2 (13) All that remains is to show that the limit in (13) is zero. In order to estimate this integral, we must rst estimate 1 1+z 2 7 on the curve R . 1 To get an upper bound for 1+z2 , we need to nd a lower bound for |1 + z 2 |. For this we use the reverse triangle inequality |z w| ||z| |w||. This gives the estimate, |1 + z 2 | |1 R2 | = R2 1, for z on R (so |z| = 1). Taking the reciprocal of this bound gives 1 1 2 1 + z2 R 1 for z on R . Since the arclength of R is R (it is a semicircle of radius R), the above bound gives an estimate for the integral: R dz 1 2 R 1 + z2 R 1 (here we used Property 6 of the line integral from your last homework). Taking the limit as R , the right side of this inequality tends to zero, so lim R R dz = 0. 1 + z2 By equation (13), we obtain dx = . 1 + x2 Exercise 5. In this exercise, we will use the Residue Theorem as in the above example to calculate dx dx. (14) x4 + 1 (a) Find the poles of 1 +1 in the upper-half plane {Im(z) > 0}, as well as their residues. f (z) = z4 (b) Calculate DR dz , z4 + 1 where DR is the half-disk in the previous example, and R is a large real number. (c) Find a constant MR such that |f (z)| MR on R . Use this bound to show that dz = 0. lim R z 4 + 1 R 8 (d) Evaluate the integral (14). Example 6. Let's use the Residue Theorem to evaluate cos(ax) dx , x2 + 1 a > 1. We would like to use the method of the previous example and integrate f (z) = cos(az) z2 + 1 (15) over the contour DR (we continue to use the notation of the last example). As in that example, we would need to show that lim R R cos(az) dz = 0. z2 + 1 (16) What does cos(az) look like on R ? From the denition of cos(z), ey + ey 2 cos(x + iy) = cos(x) + i sin(x) ey ey 2 . If we x x and let y or y , we see that | cos(z)| exponentially fast. This tells us that there is no hope for (16) to hold, as the function we are integrating is becoming very large as R . Instead of integrating f (z) from (15), let's instead try g(z) = eiaz . z2 + 1 (17) Since Re(eiax ) = cos(ax), we see that Re eiax dx = x2 + 1 cos(ax) dx . x2 + 1 (18) We will use the Residue Theorem to calculate the integral on the left. The function g(z) has simple poles at i, with Resi g(z) = so DR ea , 2i eiaz dz = a z2 + 1 e by the Residue Theorem. As in the last example, we need to show that lim R R eiaz dz = 0. z2 + 1 9 (19) We estimate, |eiaz | = |eay+iax | = eay < 1, (20) since y > 0 on R because R is in the upper-half-plane. Combining this with the estimate, 1 1 2 , 1 + z2 R 1 from the last example, we get 1 eiaz 2 1 + z2 R 1 on R . Therefore, R 1 eiaz dz 2 R 0, 2+1 z R 1 (21) (22) as R . As in the last example, we conclude that eiax dx = a x2 + 1 e since the right hand side is real, it follows from (18) that cos(ax) dx = a x2 + 1 e Exercise 6. In this exercise, we will use the Residue Theorem as in the above example to calculate cos(x) dx . (23) 2 2 (x + 1) (a) Find the poles of f (z) = cos(z) (z 2 + 1)2 in the upper-half plane {Im(z) > 0}. What are the orders of these poles? What are their residues? (b) Calculate DR cos(z) dz , (z 2 + 1)2 where DR is the half-disk in the previous example, and R is a large real number. (c) Find a constant MR such that |f (z)| MR on R . Use this bound to show that dz = 0. lim 2 + 1)2 R (z R (d) Evaluate the integral (23). 10