MAT389 Fall 2016, Problem Set 11

Transcription

1 MAT389 Fall 216, Problem Set 11 Improper integrals 11.1 In each of the following cases, establish the convergence of the given integral and calculate its value. i) x 2 x 2 + 1) 2 ii) x x 2 + 1)x 2 + 2x + 2) iii) x 3 sin x x iv) cos x x + α) 2 β > ) + β2 Remember that if the integral I = fx) converges, its value is equal to its Cauchy principal value fx) = Here is the basic strategy to compute the latter. R lim R fx). 1. Choose an appropriate complex-valued function F z) that is related to fx) on the real axis. 2. Consider the integral of F z) on the contour of Figure 1, making sure that Γ R F z) dz as R Apply the Residue theorem to conclude that F x) = 2πi j Res F z), z=z j where the sum ranges over the singularities of F z) lying on the upper half-plane. 4. Relate F x) and fx). i) At both ends of the real line, the function fx) = x 2 /x 2 + 1) 2 behaves like 1/x 2, so we expect the integral x 2 I = x 2 + 1) 2 to converge. To prove this, notice that x > x 2 over the whole real line, and so < 1 x 2 x 2 + 1) 2 < 1 x 2 x 2 ) 2 = 1 x 2 < +

2 y Γ R R R x Figure 1: Basic contour for fx) Convergence of I at the other end of the real line is shown by the exact same argument: < 1 x 2 x 2 + 1) 2 < 1 x 2 x 2 ) 2 = 1 x 2 = 1 x 2 < + Let us compute the value of the principal value of I, applying the basic strategy above: take F z) = z 2 /z 2 + 1) 2 and integrate over the contour on Figure 1. On Γ R, we have z z 2 1 = R 2 1 as long as R is greather than 1), so z ΓR 2 z 2 + 1) 2 dz R 2 R 2 πr 1) 2 F z) has two poles: one at z = i, and the other at z = i and both of them are order 2 poles. Only the first one is in the upper half-plane. The Residue theorem now gives x 2 z 2 I = x 2 = 2πi Res + 1) 2 z=i z 2 + 1) 2 = 2πi d z 2 dz z + i) 2 = 2πi i z=i 4 = π 2 ii) On the positive real axis, we have x > x 2 and x 2 + 2x + 2 > x 2, so < 1 x x 2 + 1)x 2 + 2x + 2) < 1 x x 2 x 2 = 1 x 3 < + On the negative real axis, the integrand is negative. If we switch the sign and notice that x x 2 whenever x 1 and so x 2 + 2x + 2 3x 2 there), we have < 1 x x 2 + 1)x 2 + 2x + 2) < We have proved that the integral I = 1 x x 2 3x 2 = x x 2 + 1)x 2 + 2x + 2) 1 x 3 = 1 x 3 < +

3 converges. To compute its principal value, let F z) = z/[z 2 + 1)z 2 + 2z + 2)], and use the basic contour again. On Γ R, and for R sufficiently large, we have the bounds z 2 1 R 2 1 and and hence C R z 2 + 2z + 2 = z 1 + i) z 1 i) R 2 2 z dz z 2 + 1)z 2 + 2z + 2) R R 2 1)R πr 2) 2 We can now calculate residues at the singularities of F z) lying on the upper half-plane, z = i and z = 1 + i: ) x I = x 2 + 1)x 2 + 2x + 2) = 2πi Res + Res z z=i z= 1+i z 2 + 1)z 2 + 2z + 2) ) z = 2πi z z + i)z 2 + 2z + 2) + z=i z 2 + 1)z i) ) z= 1+i 1 2i 1 + 3i = 2πi + = π iii) In this case, it is difficult to establish the convergence of the integral I = x 3 sin x x a priori, since the integrand behaves like 1/x at both ends of the real line which might indicate divergence), but its sign oscillates which always improves convergence, so maybe it converges after all). But we can use the fact that the integrand is even to show that the value of I coincides with its principal value. Indeed, the two limits that define I become one: I = lim R2 R 1,R 2 + R 1 = lim R 1 + = lim R 1 + = 2 lim On the other hand, we have R 1 R 1 R x 3 sin x x x 3 sin x x x 3 sin x x x 3 sin x x lim R2 R 2 + R2 lim R 1 + x 3 sin x R x = lim x 3 sin x R x = 2 x 3 sin x x x 3 sin x x lim R x 3 sin x x So if we compute the principal value of I and obtain a finite number, we have proven a posteriori that I converges. OK, so let us try computing that principal value. Blindly switching x s for z s that is, choosing F z) = z 3 sin z/z )) would lead to an integral over the Γ R in our basic contour Figure 1) that we cannot prove goes to zero as R +. Instead, we take F z) = z 3 e iz /z ). With

4 e ireiθ = e R sin θ, z R 4 16 this last one for R 2) and Jordan s inequality, we have z 3 e iz z dz π R4 R 4 e R sin θ dθ < R4 π 16 R 4 16 R C R On the other hand, of the four zeroes of the denominator z only two of them lie on the upper half plane those at z = 2±1 + i). By the Residue theorem, we get ) + x 3 e ix z 3 e iz x 4 = 2πi Res + 16 z= 21+i) z Res z 3 e iz z= 2 1+i) z πiz 3 e iz = z )/z 2 2πiz 3 e iz + 2i) z )/z + 2 2i) = 2πi e 2 1+i) 4 + 2πi e 2 1 i) 4 z= 21+i) = iπe 2 cos 2 z= 2 1+i) You can compute the denominator in the residue by hand, but the approach in the appendix to Problem 12.3 is faster and less error-prone). On the real line, we have x 3 e ix x 4 = + 16 x 3 cos x + i sin x) x x 3 cos x = x 4 + ii + 16 That is, we can find the integral we are interested in by taking the imaginary part: I = Im For free, we also get that x 3 e ix [ x = Im iπe 2 cos ] 2 = πe 2 cos 2 x 3 cos x x 4 =, + 16 which should not surprise given the fact that the integrand is odd. iv) In this case, the integrand behaves like a 1/x 2 at both x ±, so we can expect for I = cos x x + α) 2 + β 2 to converge. We will not dwell on proving that here. On the Complex Analysis side of things, we once again substitute the cosine for a complex exponential, taking F z) = e iz /[z +α) 2 +β 2 ]. Since z +α) 2 +β 2 = [z α + iβ)] [z α iβ)], we can bound z + α) 2 + β 2 R ) 2 α 2 + β 2 Together with Jordan s inequality, we get e iz dz z + α) 2 + β 2 R R ) 2 α 2 + β 2 C R π e R sin θ dθ < R π R ) 2 α 2 + β 2 R

5 The residue calculation yields On the real line, we have e ix x + α) 2 = 2πi Res + β2 z= α+iβ z + α) 2 + β 2 e iz = 2πi z + α + iβ = π z= α+iβ β e β iα Taking the real part, we conclude that [ e ix ] I = Re x + α) 2 + β 2 e iz e ix + x + α) 2 + β 2 = I + i sin x x + α) 2 + β 2 [ ] π = Re β e β iα = π β e β cos α The imaginary part provides another formula one that we were not looking for, but got anyway): [ sin x + x + α) 2 + β 2 = Im e ix ] x + α) 2 + β [ ] 2 π = Im β e β iα = π β e β sin α Notice that the condition β > is necessary for the singularities of the integrand to be away from the real line Compute the following integrals: i) iii) v) x 2 x 2 + 1)x 2 + 4) x x log x x 2 + 1) 2 ii) iv) vi) cos x x 2 α > ) + α2 x 5 x sin 2 x x 2 vii) x 1/4 x viii) x x 2 + 2x + 5 Hint: for vi), notice that 2 sin x 2 = 1 cos 2x = Re1 e 2ix ), and integrate the function F z) = 1 e 2iz )/z 2 over the appropriate contour. Our basic strategy for computing integrals of the type goes as follows. I = fx)

6 1. Choose a complex-valued function F z) that restricts to fx) on the real axis. 2. Choose a contour as in Figure 2a in such a way that F z) dz I, L 1 F z) dz ki L 2 k 1), F z) dz Γ R as R + ; and 3. Apply the Residue theorem to conclude that 1 + k)i = 2πi j Res F z), z=z j where the sum ranges over the singularities of F z) lying inside the contour of integration. Several variants are important: a) We might need to choose a function F z) that restricts to fx) on the positive real axis only after taking real/imaginary part. The integral over L 2 might only be proportional to I after taking real/imaginary part too. b) The function F z) might have a singularity at z =. We deal with this issue by modifying our contour of integration, carving a small circular arc about the origin as in Figure 2b. We then need to either prove that the integral over Γ ɛ also vanishes as ɛ, or compute its limiting value. c) If Θ = 2π and F z) has a branch cut along the positive real axis, we use the keyhole contour of Figure 2c and let δ. i) We choose F z) = z 2 z 2 + 1)z 2 + 4) Since this function is even, we take Θ = π, in which case we obtain k = 1. Over Γ R for R sufficiently large, we have z 2 = R 2, z 2 +1 R 2 1 = R 2 1 and z 2 +4 R 2 4 = R 2 4. Hence, z ΓR 2 z 2 + 1)z 2 + 4) dz R 2 R 2 1)R 2 πr 4) and z 2 2I = 2πi Res z=i z 2 + 1)z 2 + 4) + Res z 2 ) z=2i z 2 + 1)z 2 + 4) z 2 = 2πi z 2 ) z + i)z 2 + 4) + z=i z 2 + 1)z + 2i) z=2i i = 2πi 6 i ) = π = I = π ii) We use variant a) of the basic strategy, with F z) = e iz /z 2 + α 2 ) and Θ = π. Bounding z 2 + α 2 R 2 α 2 on Γ R for suficciently large R, and using Jordan s inequality, we have e CR iz z 2 + α 2 dz R π R α) 2 e R sin θ R π dθ < R α) 2 R

7 y y Γ R Γ R Re iθ Re iθ L 2 Θ L 2 Γ ɛ Θ L 1 R x L 1 R x a) The basic contour b) Avoiding the origin y Γ R Γ ɛ L 1 ɛeiδ ɛe iδ L 2 Reiδ δ δ Re iδ x c) The case Θ = 2π with a branch cut Figure 2: Contours for fx) The integration over the real line yields e ix x 2 + α 2 + e ix x 2 + α 2 = e ix + e ix x 2 + α 2 = 2I We can now apply to Residue theorem to get e ix 2I = 2πi Res z=αi x 2 + α 2 = 2πi e ix x + αi = 2πi e α z=αi 2αi = π α e α = I = π 2α e α

8 iii) Let F z) = z/z 5 + 1). Under the transformation z e 2πi/5 z, the function F z) gets multiplied by the constant factor e 2πi/5. Thus we can apply the basic strategy with Θ = 2π/5, which results in k = e 4πi/5 ; indeed, L 2 z z dz = e 2πi/5 x R R x e2πi/5 = e 4πi/5 x x e4πi/5 I Over Γ R, we have z = R, z R 5 1 = R 5 1 and z z dz R 2π R R C R The denominator in F z) has only one singularity inside the contour of integration: a simple pole at z = e πi/5. Hence, 1 e 4πi/5 z )I = 2πi Res z=e πi/5 z = 2πi z z 5 + 1)/z e πi/5 ) = 2πi 1 z=e πi/5 5 e 3πi/5 π = I = 5 sin2π/5) For the calculation of the denominator in the residue, see the appendix to this problem). iv) Let F z) = z 5 /z 1 + 1). Since the denominator is even, we might try Θ = π; however, that produces k = 1 after all, the original fx) is odd). Following on the example of the last integral, we take Θ = 2π/1 = π/5 instead and find that, with this choice, k = e πi/5. It is easy to see that the integral on Γ R vanishes in the limit: z dz ΓR z 5 R5 π R R Only one of the poles of F z) lie inside of the contour of integration, namely z = e πi/1. Hence, 1 + e πi/5 z 5 )I = 2πi Res z=e πi/1 z = 2πi z 5 z 1 + 1)/z e πi/1 ) = 2πi 1 z=e πi/1 1 e 2πi/5 π = I = 1 cosπ/1) Once again, refer to the appendix for the computation of the denominator in the residue). v) Take F z) = log ) z/z 2 + 1) 2. Since the denominator stays the same after z z, we can take Θ = π. Two difficulties arise. The first one is that F z) has a branch point at the origin, and so we will need to invoke variant c) of our basic strategy. Moreover, the integral over L 2 is proportional to that over L 1 only after taking real parts that is, we need to blend in variant b) as well. Let us start by proving that the integrals over Γ R and Γ ɛ are zero in the limits R + and ɛ, respectively. For the first, and with R sufficiently large, we have log ) z = Log R + iθ Log R + π, z R 2 1 = R 2 1 log ) z Γ R z 2 + 1) 2 dz Log R + π R 2 πr 1) 2

9 On the small circular arc of radius ɛ < 1, it is log ) z = Log ɛ + iθ Log ɛ + π, z ɛ 2 1 = 1 ɛ 2 log ) z Γ ɛ z 2 + 1) 2 dz Log ɛ + π 1 ɛ 2 ) 2 πɛ ɛ Over the straight legs of the contour, we get log ) z R L 1 z 2 + 1) 2 dz = Log x ɛ x 2 I + 1) 2 log ) z ɛ L 2 z 2 + 1) 2 dz = Log x) + iπ + x 2 + 1) 2 I + iπ x 2 + 1) 2 R as both R + and ɛ. The point z = i is a double pole of fz), so log ) z 2I + iπ x 2 = 2πi Res + 1) 2 z=i z 2 + 1) 2 = 2πi d log ) z dz z + i) 2 = 2πi z + i 2z log ) z zz + i) 3 = π z=i 2 + π2 4 i Taking the real part of this last expression yields the integral we seek: z=i As a bonus, we obtain I = π 4 x 2 + 1) 2 = π 4 from the imaginary part vi) As per the hint, take F z) = 1 e 2iz )/z 2. Notice that this function has a simple pole at the origin, and so we will need to use variant c) of the basic strategy. Since fx) is even, we try Θ = π. This choice yields, in the limit, L1 1 e 2iz z 2 = R ɛ 1 e 2ix x 2 sin 2x 2I i x 2 1 e L2 2iz ɛ 1 e 2ix R 1 e 2ix sin 2x z 2 = R x 2 = ɛ x 2 2I + i x 2 The integral over Γ R vanishes: 1 e ΓR 2iz z 2 dz 2 πr R2 That over Γ ɛ, however, does not: Cɛ 1 e 2iz z 2 1 e 2iz dz = πi Res z= z 2 = 2π Since F z) has no poles inside the contour of integration, we have 4I 2π =, or sin 2 x x 2 = π 2

10 vii) Let F z) = z 1/4 /z 3 + 1), where the numerator is the determination z 1/4 = e 1/4) log ) z that has the branch cut along the negative imaginary axis so we will need an indented contour like that of Figure 2b and restricts to x 1/4 on the positive real axis. Since the denominator is invariant under the transformation z e 2πi/3 z, we set Θ = 2π/3, which gives L2 z 1/4 z dz = ɛ R L1 z 1/4 z dz = e πi/6 x 1/4 x R ɛ x 1/4 x 3 I + 1 R e2πi/3 = e 5πi/6 x 1/4 x e5πi/6 I On the other hand, the integrals over Γ R and Γ ɛ go to zero as R + and ɛ : z CR 1/4 z dz R1/4 2π R R z dz ɛ1/4 2π 1 ɛ 3 3 ɛ Cɛ z 1/4 ɛ ɛ Only the pole at z = e πi/3 is inside the contour of integration, so 1 e 5πi/6 z 1/4 )I = 2πi Res z=e πi/3 z = 2πi z 1/4 z 3 + 1)/z e πi/3 ) π = I = 3 sin5π/12) z=e πi/3 = 2πi 1 3 e 7πi/12 The denominator can be calculated by hand, but the argument of the appendix still applies). viii) Here we need to resort to variant d) of the basic strategy, since the polynomial z 2 + 2z + 5 is not invariant under any transformation of the form z e iθ z with Θ 2πZ. Thus, we let F z) = z 1/2 /z 2 + 2z + 5), where by z 1/2 we mean the determination z 1/2 = e 1/2) log ) z that has its branch cut along the positive real axis. On the two legs of the contour, we have L 2 L 1 z 1/2 dz R z 2 + 2z + 5 = e iδ/2 x 1/2 e iδ ɛ e 2iδ x 2 + 2e iδ x + 5 I z 1/2 dz ɛ z 2 + 2z + 5 = e i2π δ)/2 x 1/2 e i2π δ) R e 2i2π δ) x 2 + 2e i2π δ) x + 5 I where the limit is as R +, ɛ and δ. Over the two circular arcs, z 1/2 dz z 2 + 2z + 5 R 1/2 R 2π 2δ)R 5) 2 Γ R Γ ɛ where we have used the bounds z 1/2 dz z 2 + 2z + 5 ɛ 1/2 2π 2δ)ɛ 5 ɛ) 2 ɛ z 2 + 2z + 5 = z + 1 2i z i R 5) 2

11 z 2 + 2z + 5 = z + 1 2i z i 5 ɛ) 2 Hence, z 1/2 2I = 2πi Res z= 1+2i z 2 + 2z Res z= 1 2i z 2 + 2z + 5 ) z 1/2 z 1/2 = 2πi + z i z + 1 2i z= 1+2i ) z= 1 2i 1 + 2i) 1/2 1 2i)1/2 = 2πi = π [ 1 + 2i) 1/ i) 1/2] 4i 4i 2 = I = π [ 1 + 2i) 1/ i) 1/2] = π ) 4 4 z 1/2 ) Appendix: in iii), iv) and vii), the following quantity appeared when calculating the appropriate residue: z k k z e πi/k = z e 2j+1)πi/k) z=e πi/k j=1 Although a brute force computation is possible for low values of k, it quickly becomes unmanageable. A slick way of computing it for all values of k at once! is to apply l Hôpital s rule, which we proved in Problem 11.2 the result in Problem is, in fact, enough : z k + 1 z k + 1 z e πi/k = lim z=e πi/k z e πi/k z e πi/k = lim kz k 1 z e πi/k 1 = kzk 1 1 z=e πi/k = ke k 1)πi/k 11.3 Derive the integration formula e x2 cos 2bx = π 2 e b2 by integrating the function fz) = e z2 around the rectangular contour C in the figure, and then letting a +. Use the well-known integration formula a + bi e x2 = y π 2. a + bi a a x

12 First of all, note that, since fz) is entire, C fz)dz =. Hence the integrals over each of the segments in the contour with the orientation given) sum up to zero. Label them as follows: C 1, along the real axis; C 2, the vertical segment on the right; C 2, the top of the contour; and C 4, the vertical segment on the left. We claim that the integrals over C 2 and C 4 vanish in the limit a +. Indeed, on both of them we have e z2 = e a2 +y 2 e a2 +b 2, so dz +b 2 e a2 b and dz +b 2 e a2 b. C 2 e z2 C 4 e z2 On the real axis we get the well-know Gaussian integral: a dz = e x2 e x2 = 2 a + C 1 e z2 The integral on C 4 yields dz = C 4 e z2 a a a e x2 = π a e x2 +b 2 2ibx = e b2 e x2 cos 2bx i sin 2bx) a + eb2 e x2 cos 2bx + ie b2 e x2 sin 2bx From the real part of C fz)dz = we get a e x2 cos 2bx = πe b2 = since the integrand is an even function. e x2 cos 2bx = π 2 e b2, 11.4 Show that cos x π e x = + e x 2 cosh) Hint: integrate fz) = e iz /e z + e z ) around the rectangular contour C in the figure, and then let R +. R + iπ y R + iπ R R x Let C denote the integration contour defined in the hint. We refer to the line segments that make it up as follows:

13 C 1, from R to R along the real axis; C 2, the vertical segment from R to R + iπ; C 2, the top of the contour from R + π to R + iπ; and C 4, the vertical segment from R + iπ to R. The points of C 2 are of the form z = R + iy with y π. Hence the following bounds hold there: We can thus show that e iz = e ir+iy) = e y 1, e z + e z e z e z = e R e R = sinh R. C 2 e iz dz e z + e z We can treat the integral along C 4 similarly: π sinh R e iz = e i R+iy) = e y 1, e z + e z e z e z = e R e R = sinh R; e iz dz e z + e z π sinh R C 4 Along C 1 we have e iz dz R C 1 e z + e z = e ix e ix R e x + e x e x + e x The integral we are interested in is the real part of the latter. The integral on C 3 is closely related to it too: C3 e iz e z dz = + e z R R e ix+iπ) R e x+iπ = e π + e x iπ R e ix e x + e x e π e ix e x + e x For the residue calculation, notice that e z + e z = 2 cosh z = for z = 2k + 1)πi/2 with k Z. Of these, only one is inside the contour of integration when k =. Using our knowledge of Taylor series, we have Res z=πi/2 Putting it all together, e iz e z = Res + e z z=πi/2 1 + e π ) e iz 2 cosh z = Res z=πi/2 2i n= z πi/2)2n+1 /2n + 1)! 1 e iz = Res z=πi/2 z πi/2 2i n= z πi/2)2n /2n + 1)! e iz = 2i n= z πi/2)2n /2n + 1)! = e z=πi/2 2i e ix e x = 2πi + e x Res = z=πi/2 e iz e iz e z = πe + e z e ix e x + e x = πe 1 + e π ) = π cosh) The real part of this identity is the result we were looking for and the imaginary part of the integral on the left is zero.

14 Laplace transforms 11.5 Find the inverse Laplace transforms of the following functions: i) F s) = 1 ii) F s) = 3 5s 2s 2 s + 1)s 2 + 2s + 5) iii) F s) = 2s3 s 4 F s) = s2 a 2 iv) 4 s 2 + a 2 ) 2 Let F s) be a function that is holomorphic everywhere on the complex plane except for a finite collection of isolated singularities, z k k = 1,..., n). Recall that the inverse Laplace transform of F s) is defined as L 1 {F s)}t) = 1 γ+i 2πi e st F s) ds, where γ is a any real number satisfying γ > Re z k for all k. We calculate this by applying the Residue theorem to the contour in Figure 3. If we can prove that the integral over Γ R vanishes as R +, we can conclude that n L 1 [ {F s)}t) = Res e st F s) ]. z=z k y k=1 γ i γ + ir Γ R γ x γ ir Figure 3: The contour for the inverse Laplace transform i) The function given has just one simple) pole at s = 3/5, and so we fix γ > 3/5. Parametrizing Γ R as s = γ + Re iθ for < θ < 3, we have e ΓR st 3 5s ds 3 = e tγ+reiθ ) 3 ire iθ e tγ+reiθ ) dθ 3 5s 3 5γ 5Re iθ R dθ

15 As long as R is sufficiently big, we can bound the denominator from below by 3 5γ 5Re iθ 3 5γ 5R = 5R 5γ + 3 On the other hand, the numerator can be written as e tγ+reiθ ) = e γt e Rt cos θ With the change of variables θ = ψ +, Jordan s inequality yields e CR st 3 5s ds 3 Re γt ert cos θ 5R 5γ + 3 dθ = Re γt π e Rt sin ψ dψ < 5R 5γ + 3 The latter goes to zero as R +, and so L 1 e st {F s)}t) = Res s=3/5 3 5s = 1 5 e3t/5 Re γt 5R 5γ + 3 ii) Since s 2 + 2s + 5 = [ s 1 + 2i) ][ s 1 2i) ], the function F s) has three poles: s = 1, s = 1 + 2i and s = 1 2i. We thus need to take γ > 1. For simplicity, me may choose γ =. As above, we parametrize Γ R by s = γ + Re iθ = Re iθ for < θ < 3, so that 2s 2)e ΓR st s + 1)s 2 + 2s + 5) ds 3 2Re iθ 1)e treiθ = Re iθ + 1) [ Re iθ 1 + 2i) ][ Re iθ 1 2i) ] ireiθ dθ 3 2R Re iθ 1 e tre iθ Re iθ + 1 Re iθ + 1 2i Re iθ i dθ In the numerator of this last integrand, we have Re iθ 1 R + 1 e treiθ = e Rt cos θ In the denominator, we bound Re iθ + 1 R 1 = R 1 Re iθ + 1 2i R 1 2i = R 5 Re iθ i R 1 + 2i = R 5 as long as R is big enough. Putting this all together, performing the change of variables θ = ψ + and using Jordan s inequality, we have 2s 2)e ΓR st s + 1)s 2 + 2s + 5) ds 3 2RR + 1)eRt cos θ R 1)R 5) dθ 2 π 2RR + 1) = R 1)R e Rt sin ψ dψ 5) 2 2RR + 1) < R 1)R π 5) 2 Rt π Rt

16 Since this goes to zero in the limit, we conclude that L 1 {F s)}t) = Res Res Res + + s= 1 s= 1+2i s= 1 2i = e t + 1 i e t e 2it i e t e 2it 2 2 = e t[ cos 2t + sin 2t 1 ] ) 2s 2)e st s + 1)s 2 + 2s + 5) iii) We have four poles, located at the points s = ± 2 and s = ± 2i, so we fix γ > 2. Calculating as above, we have 2s ΓR 3 e st s 4 4 ds 3 = 2γ + Re iθ ) 3 e tγ+reiθ ) γ + Re iθ ) 4 ire iθ dθ 4 3 2R γ + Re iθ 3 e tγ+re iθ ) γ + Re iθ ) 4 dθ 4 In the numerator, we write γ + Re iθ e tγ+reiθ ) R + γ = e γt e Rt cos θ For R sufficiently big, we have γ + Re iθ R γ and γ + Re iθ ) 4 4 γ + Re iθ 4 4 = γ + Re iθ 4 4 R γ) 4 4 Once again we use this information, the change of variables θ = ψ + and Jordan s inequality to obtain 2s ΓR 3 e st s 4 4 ds 3 2RR + γ) 3 e γt ert cos θ R γ) 4 dθ 4 = 2RR + γ)3 e γt R γ) 4 4 < 2RR + γ)3 e γt R γ) 4 4 π e Rt sin ψ dψ π Rt The residue computation is tedious but straightforward. It yields ) 2s L 1 3 e st {F s)}t) = Res s= + Res 2 s= + Res 2 s= + Res 2i s= 2i s 4 4 = 1 2 e 2t e 2t ei 2t e i 2t = cosh 2t + cos 2t iv) F s) has double poles at s = ±ai, so we need γ >. Then, s ΓR 2 a 2 )e st 3 s 2 + a 2 ) 2 = γ + Re iθ a)γ + Re iθ + a)e tγ+reiθ ) γ + Re iθ ai) 2 γ + Re iθ + ai) 2 ire iθ dθ 3 R γ + Re iθ a γ + Re iθ + a e tγ+re iθ ) γ + Re iθ ai 2 γ + Re iθ + ai 2 dθ

17 For the bounds, we take γ + Re iθ a R + γ a γ + Re iθ + a R + γ + a e tγ+reiθ ) = e γt Rt cos θ e γ + Re iθ ai R γ ai = R γ ai γ + Re iθ + ai R γ + ai = R γ + ai as long as R is big enough. With these, s ΓR 2 a 2 )e st 3 s 2 + a 2 ) 2 = < RR + γ a )R + γ + a )e γt ert cos θ R γ ai ) 2 R γ + ai ) 2 dθ π RR + γ a )R + γ + a )eγt R γ ai ) 2 R γ + ai ) 2 e Rt sin ψ dψ RR + γ a )R + γ + a )eγt π R γ ai ) 2 R γ + ai ) 2 Rt We can thus evaluate the inverse Laplace transform by calculating residues: L 1 s 2 a 2 )e st {F s)}t) = Res s=ai s 2 + a 2 ) 2 + Res s= ai = d s 2 a 2 )e st dz s + ai) 2 + d s=ai dz = 1 2 teiat te iat = t cos at s 2 a 2 )e st s 2 + a 2 ) 2 s 2 a 2 )e st s ai) 2 s= ai 11.6 Using Laplace transforms, solve the following initial value problems: i) y + y = sin 4t, y) =, y ) = 1, ii) y + y + 2y = e t cos 2t, y) = 1, y ) = 1, Recall the Laplace transform of sines and cosines, L{sin ωt} = and the following basic properties, where F s) = Lft)s). ω s 2 + ω 2, L{cos ωt} = s s 2 + ω 2, L{f t)}s) = sf s) f), L{f t)}s) = s 2 F s) sf) f ) L{e ct ft)}s) = F s c)

18 i) Thanks to the formulas above and the linearity of the Laplace transform, we have L{y t)}s) + L{yt)}s) = L{sin 4t}s) s 2 4 Y s) 1 + Y s) = s s Y s) = s 2 + 1)s ) Notice that Y s) has poles at s = ±i and s = ±4i. Hence, for γ >, we write s ΓR 2 + 2)e st s 2 + 1)s ) ds 3 [ = γ + Re iθ ) ] e tγ+reiθ ) [ γ + Re iθ ) ][ γ + Re iθ ) ] ireiθ dθ 3 R γ + Re iθ ) e tγ+re iθ ) γ + Re iθ ) γ + Re iθ ) dθ For large R, we calculate the following bounds: γ + Re iθ ) = γ + Re iθ + i 2 γ + Re iθ i 2 R + γ + i ) 2 R + γ i ) 2 e tγ+reiθ ) = e γt e Rt cos θ γ + Re iθ ) = γ + Re iθ + i γ + Re iθ i R + γ + ) 2 2 R γ + i ) R γ i ) R γ 1 ) 2 γ + Re iθ ) = γ + Re iθ + 4i γ + Re iθ 4i R γ + 4i ) R γ 4i ) R γ 4 ) 2 Then, s ΓR 2 + 2)e st s 2 + 1)s ) ds 3 RR + γ + 2) 2 e γt ert cos θ R γ 1 ) 2 R γ 4 ) 2 dθ = < RR + γ + 2) 2 e γt R γ 1 ) 2 R γ 4 ) 2 π e Rt sin ψ dψ RR + γ + 2) 2 e γt π R γ 1 ) 2 R γ 4 ) 2 Rt We can thus calculate the inverse Laplace transform by evaluating residues: ) s yt) = L 1 2 {Y s)}t) = Res + Res + Res + Res + 2)e st z=i z= i z=4i z= 4i s 2 + 1)s ) = 19 3i eit 19 3i e it 1 3i e4it + 1 3i e 4it = 1 [ ] 19 sin t sin 4t 15

19 ii) [This problem is computationally heavy. Don t worry too much about the explicit calculation of the residues: I used a computer to do it!] The Laplace transform of the given initial value problem yields L{y t)}s) + L{y t)}s) + 2L{yt)}s) = L{e t cos 2t}s) s 2 s + 1 Y s) s sy s) 1 + 2Y s) = s + 1) The poles of this function are located at Y s) = s 3 + 2s 2 + 6s + 1 s 2 + 2s + 5)s 2 + s + 2) s 2 + 2s + 5 = s = 1 ± 2i, and s 2 + s + 2 = s = 1 2 ± 7 2 i We thus need γ > 1/2. It will be convenient for our calculations below to set γ =. Then, s ΓR 3 + 2s 2 + 6s + 1)e st s 2 + 2s + 5)s 2 + s + 2) ds 3 R 3 e 3iθ + 2R 2 e 2iθ + 6Re iθ + 1 ) e treiθ = R 2 e 2iθ + 2Re iθ + 5) R 2 e 2iθ + Re iθ + 2) ireiθ dθ 3 R R 3 e 3iθ + 2R 2 e 2iθ + 6Re iθ + 1 e tre iθ R 2 e 2iθ + 2Re iθ + 5 R 2 e 2iθ + Re iθ + 2 dθ Although this integral might seem a bit daunting, we can work out each piece simply: R 3 e 3iθ + 2R 2 e 2iθ + 6Re iθ + 1 R 3 + 2R 2 + 6R + 1 e treiθ = e Rt cos θ R 2 e 2iθ + 2Re iθ + 5 = R i R + 1 2i R 2 e 2iθ + Re iθ + 2 = R i R i Once again, we have assumed above that R is sufficiently big. s ΓR 3 + 2s 2 + 6s + 1)e st s 2 + 2s + 5)s 2 + s + 2) ds 3 R ) 2 5 R ) 2 2 RR 3 + 2R 2 + 6R + 1)eRt cos θ ) 2 ) 2 dθ R 5 R 2 = RR3 + 2R 2 + 6R + 1) π ) 2 ) 2 R 5 R 2 e Rt sin ψ dψ < RR3 + 2R 2 + 6R + 1) π ) 2 ) 2 R 5 R 2 Rt

20 The residue calculation yields yt) = L 1 {Y s)}t) = Res + Res + z= 1+2i z= 1 2i = 1 + i 8 e 1+2i)t 1 + i 8 Res z= 1+ 7i)/2 = 1 4 e t[ cos 2t + sin 2t ] e t/2 [5 cos + Res z= 1 7i)/2 ) s 3 + 2s 2 + 6s + 1)e st s 2 + 2s + 5)s 2 + s + 2) e 1 2i)t i 5 7i)t/ i e i 5 7 7i e 1 ] t 7 sin 2 t 7i)t/2