Topic 9 Notes Jeremy Orloff 9 Definite integrals using the residue theorem 9.1 Introduction In this topic we'll use the residue theorem to compute some real definite integrals. Z b f(x) dx a The general approach is always the same 1. Find a complex analytic function g(z) which either equals f on the real axis or which is closely connected to f, e.g. f(x) = cos(x), g(z) = eiz. 2. Pick a closed contour C that includes the part of the real axis in the integral. 3. The contour will be made up of pieces. It should be such that we can compute Z g(z) dz over each of the pieces except the part on the real axis. Z 4. Use the residue theorem to compute g(z) dz. C 5. Combine the previous steps to deduce the value of the integral we want. 9.2 Integrals of functions that decay The theorems in this section will guide us in choosing the closed contour C described in the introduction. The first theorem is for functions that decay faster than 1=z. Theorem 9.1. (a) Suppose f(z) is defined in the upper half-plane. If there is an a > 1 and M > 0 such that M f(z) < j j z a j j for z large then j j Z lim f(z) dz = 0; R!1 CR where CR is the semicircle shown below on the left. Im(z) Im(z) R R − Re(z) CR CR Re(z) R R − 1 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 2 Semicircles: left: Reiθ, 0 < θ < π right: Reiθ, π < θ < 2π. (b) If f(z) is defined in the lower half-plane and M f(z) < ; j j z a j j where a > 1 then Z lim f(z) dz = 0; R!1 CR where CR is the semicircle shown above on the right. Proof. We prove (a), (b) is essentially the same. We use the triangle inequality for integrals and the estimate given in the hypothesis. For R large Z Z Z Z π M M Mπ f(z) dz f(z) dz dz = R dθ = : ≤ j j j j ≤ z a j j Ra Ra−1 CR CR CR j j 0 Since a > 1 this clearly goes to 0 as R . QED ! 1 The next theorem is for functions that decay like 1=z. It requires some more care to state and prove. Theorem 9.2. (a) Suppose f(z) is defined in the upper half-plane. If there is an M > 0 such that M f(z) < j j z j j for z large then for a > 0 j j Z lim f(z)eiaz dz = 0; x1!1; x2!1 C1+C2+C3 where C1 + C2 + C3 is the rectangular path shown below on the left. Im(z) Im(z) C x2 x1 2 − Re(z) i(x1 + x2) C3 C1 C3 C1 Re(z) x x1 i(x1 + x2) C2 − 2 − Rectangular paths of height and width x1 + x2. (b) Similarly, if a < 0 then Z lim f(z)eiaz dz = 0; x1!1; x2!1 C1+C2+C3 where C1 + C2 + C3 is the rectangular path shown above on the right. Note. In contrast to Theorem 9.1 this theorem needs to include the factor eiaz. Proof. (a) We start by parametrizing C1, C2, C3. C1: γ1(t) = x1 + it, t from 0 to x1 + x2 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 3 C2: γ2(t) = t + i(x1 + x2), t from x1 to x2 − C3: γ3(t) = x2 + it, t from x1 + x2 to 0. − Next we look at each integral in turn. We assume x1 and x2 are large enough that M f(z) < j j z j j on each of the curves Cj. Z Z Z iaz iaz M iaz f(z)e dz f(z)e dz e dz ≤ j j j j ≤ z j j j j C1 C1 C1 j j Z x1+x2 M = eiax1−at dt p 2 2 0 x1 + t j j M Z x1+x2 e−at dt ≤ x1 0 M = (1 e−a(x1+x2))=a: x1 − Since a > 0, it is clear that this last expression goes to 0 as x1 and x2 go to . 1 Z Z Z iaz iaz M iaz f(z)e dz f(z)e dz e dz ≤ j j j j ≤ z j j j j C2 C2 C2 j j Z x1 M = eiat−a(x1+x2) dt p 2 2 −x2 t + (x1 + x2) j j Me−a(x1+x2) Z x1+x2 dt ≤ x1 + x2 0 Me−a(x1+x2) ≤ Again, clearly this last expression goes to 0 as x1 and x2 go to . 1 The argument for C3 is essentially the same as for C1, so we leave it to the reader. The proof for part (b) is the same. You need to keep track of the sign in the exponentials and make sure it is negative. Example. See Example 9.16 below for an example using Theorem 9.2. Z 1 Z 1 9.3 Integrals and −∞ 0 Example 9.3. Compute Z 1 1 I = 2 2 dx: −∞ (1 + x ) Solution: Let f(z) = 1=(1 + z2)2: 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 4 It is clear that for z large f(z) 1=z4: ≈ In particular, the hypothesis of Theorem 9.1 is satisfied. Using the contour shown below we have, by the residue theorem, Z X f(z) dz = 2πi residues of f inside the contour. (1) C1+CR Im(z) CR i Re(z) R C R − 1 We examine each of the pieces in the above equation. Z f(z) dz: By Theorem 9.1(a), CR Z lim f(z) dz = 0: R!1 CR Z f(z) dz: Directly, we see that C1 Z Z R Z 1 lim f(z) dz = lim f(x) dx = f(x) dx = I: R!1 R!1 C1 −R −∞ So letting R , Equation 1 becomes ! 1 Z 1 X I = f(x) dx = 2πi residues of f inside the contour. −∞ Finally, we compute the needed residues: f(z) has poles of order 2 at i. Only z = i is ± inside the contour, so we compute the residue there. Let 1 g(z) = (z i)2f(z) = : − (z + i)2 Then 2 1 Res(f; i) = g0(i) = = −(2i)3 4i So, π I = 2πi Res(f; i) = : 2 Example 9.4. Compute Z 1 1 I = 4 dx: −∞ x + 1 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 5 Solution: Let f(z) = 1=(1 + z4). We use the same contour as in the previous example Im(z) CR ei3π/4 eiπ/4 Re(z) R C R − 1 As in the previous example, Z lim f(z) dz = 0 R!1 CR and Z Z 1 lim f(z) dz = f(x) dx = I: R!1 C1 −∞ So, by the residue theorem Z X I = lim f(z) dz = 2πi residues of f inside the contour. R!1 C1+CR The poles of f are all simple and at eiπ=4; ei3π=4; ei5π=4; ei7π=4: Only eiπ=4 and ei3π=4 are inside the contour. We compute their residues as limits using iπ=4 L'Hospital's rule. For z1 = e : −i3π=4 z z1 1 1 e Res(f; z1) = lim (z z1)f(z) = lim − 4 = lim 3 = i3π=4 = z!z1 − z!z1 1 + z z!z1 4z 4e 4 i3π=4 and for z2 = e : −iπ=4 z z2 1 1 e Res(f; z2) = lim (z z2)f(z) = lim − 4 = lim 3 = i9π=4 = z!z2 − z!z2 1 + z z!z2 4z 4e 4 So, 1 i 1 i 2i p2 I = 2πi(Res(f; z1) + Res(f; z2)) = 2πi − − + − = 2πi = π 4p2 4p2 −4p2 2 Example 9.5. Suppose b > 0. Show Z 1 cos(x) πe−b 2 2 dx = : 0 x + b 2b Solution: The first thing to note is that the integrand is even, so 1 Z 1 cos(x) I = 2 2 : 2 −∞ x + b 9 DEFINITE INTEGRALS USING THE RESIDUE THEOREM 6 Also note that the square in the denominator tells us the integral is absolutely convergent. We have to be careful because cos(z) goes to infinity in either half-plane, so the hypotheses of Theorem 9.1 are not satisfied. The trick is to replace cos(x) by eix, so Z 1 ix ~ e 1 ~ I = 2 2 dx; with I = Re(I): −∞ x + b 2 Now let eiz f(z) = : z2 + b2 For z = x + iy with y > 0 we have ei(x+iy) e−y f(z) = j j = : j j z2 + b2 z2 + b2 j j j j Since e−y < 1, f(z) satisfies the hypotheses of Theorem 9.1 in the upper half-plane. Now we can use the same contour as in the previous examples Im(z) CR ib Re(z) R C R − 1 We have Z lim f(z) dz = 0 R!1 CR and Z Z 1 lim f(z) dz = f(x) dx = I:~ R!1 C1 −∞ So, by the residue theorem Z X I~ = lim f(z) dz = 2πi residues of f inside the contour. R!1 C1+CR The poles of f are at bi and both are simple. Only bi is inside the contour. We compute ± the residue as a limit using L'Hospital's rule eiz e−b Res(f; bi) = lim (z bi) = : z!bi − z2 + b2 2bi So, πe−b I~ = 2πi Res(f; bi) = : b Finally, 1 πe−b I = Re(I~) = ; 2 2b as claimed.