Theorems on Complex Integrals and Exponentials

Complex Integrals and the Laplace Transform There are a few computations for which the complex exponential is very nice to use. We’ll see a few here, but first a couple of Theorems about integrating a complex function: Theorem: Proof: Z Z e(bi)t dt = e(bi)t dt = Z 1 (bi)t e bi e(bt)i dt = Z cos(bt) + i sin(bt) dt = Z cos(bt) dt + i Z sin(bt) dt = 1 i sin(bt) − i cos(bt) sin(bt) − cos(bt) = b b b And 1 (bt)i cos(bt) + i sin(bt) i − sin(bt) + i cos(bt) sin(bt) − i cos(bt) e = · = = bi bi i −b b Therefore, these quantities are the same. 1 e(a+bi)t (a + bi) You can work this out, but it is more complicated since we’ll need to do integration by parts twice for each integral. It is a nice exercise to try out when you have a little time. Theorem: Z e(a+bi)t dt = Theorem: The main computational technique is using the following: 1 e(a+ib)t e cos(bt) dt = Re e dt = Re a + ib Z    Z 1 (a+ib)t at (a+bi)t e e sin(bt) dt = Im e dt = Im a + ib Z Z at (a+bi)t    Worked Example: 1. Use complex exponentials to compute Z e2t cos(3t) dt. SOLUTION: We note that e2t cos(3t) = Re(e(2+3i)t ), so: 1 e(2+3i)t 2 + 3i Simplifying the term inside the parentheses and multiplying out the complex terms: Z e2t cos(3t) dt = Re e2t e 2t     2 − 3i (cos(3t) + i sin(3t)) = 4+9  2 3 3 2 cos(3t) + sin(3t) + i − cos(3t) + sin(3t) 13 13 13 13    Therefore, Z e2t cos(3t) dt = e2t  1 3 2 cos(3t) + sin(3t) 13 13  In fact, we get the other integral for free: Z 2t 2t  e sin(3t) dt = e 2. Use complex exponentials to compute Z −3 2 cos(3t) + sin(3t) 13 13  sin(at) dt This one is simple enough to do without using complex exponentials, but it does still work.   Z  Z 1 (at)i e dt = Im sin(at) dt = Im (cos(at) + i sin(at) = ai      1 −1 −1 −i (cos(at) + i sin(at)) = Im sin(at) + i cos(at) = cos(at) Im a a a a 3. Use complex exponentials to compute the Laplace transform of cos(at): SOLUTION: Note that cos(at) = Re(e(at)i ) L(cos(at)) = Z ∞ e −st Z ∞ Z ∞ e cos(at) dt = Re e  dt = 0 0 Re −st (ai)t  e−(s−ai)t dt = Re 0 −1 e−(s−ai)t (s − ai) t→∞ t=0 What happens to our expression as t → ∞? The easiest way to take the limit is to check the magnitude (see if it is going to zero): −1 −st (ai)t −1 = · e−st · e(ai)t e e s − ai s − ai Now, the first term is a constant and e(at)i is a point on the unit circle (so its magnitude is 1). Therefore, the magnitude depends solely on e−st , where s is any real number. And, the function e−st → 0 as t → ∞ for any s > 0. Therefore, −1 e−(s−ai)t = 0 t→∞ (s − ai) lim and the Laplace transform is:  L(cos(at)) = Re 0 − −1 s + ai = Re 2 s − ai s + a2    = s2 s + a2 As a side remark, we get the Laplace transform of sin(at) for free since it is the imaginary part. 2 Homework Addition to Section 6.1 1. Use complex exponentials to compute Z e−2t sin(3t) dt. 2. Use complex exponentials to compute the Laplace transform of sin(at). 3. Use complex exponentials to compute the Laplace transform of eat sin(bt) and eat cos(bt) (compare to exercises 13, 14). 4. Show that, if f (t) is bounded (that is, there is a constant A so that |f (t)| ≤ A for all t), then f is of exponential order (do this by finding K, a and M from the definition). 5. If the function is of exponential order, find the K, a and M from the definition. Otherwise, state that it is not of exponential order. Something that may be handy from algebra: A = eln(A) . 2 (a) sin(t) (d) et (b) tan(t) (e) 5t (c) t3 (f) tt 6. Use complex exponentials to find the Laplace transform of t sin(at). 3 Homework Addition Solutions 1. Use complex exponentials to compute Z e−2t sin(3t) dt. SOLUTION: Z  Im e −2t sin(3t) dt = Z  (−2+3i)t Im e 1 dt = Im e(−2+3i)t = −2 + 3i    2 3 3 2 − − i · (e−2t cos(3t) + ie−2t sin(3t) = − e−2t cos(3t) − e−2t sin(3t) 13 13 13 13   2. Use complex exponentials to compute the Laplace transform of sin(at). SOLUTION: L(sin(at)) = Z ∞ e−st sin(at) dt 0 Ignoring the bounds for a bit, Z Im e (−s+ai)t 1 dt = Im e(−s+ai)t = −s + ai    a s − 2 i · (e(−s+ai)t Im − 2 2 s +a s + a2 As we showed earlier, if t → ∞, then   t→∞ t=0 e−(s−ai)t = e−st e(at)i → 0 as long as s > 0 (because |e(at)i | = 1). Therefore, L(sin(at)) = 0 − − s2 a a = 2 2 +a s + a2 3. Use complex exponentials to compute the Laplace transform of eat sin(bt) and eat cos(bt) (compare to exercises 13, 14). SOLUTION: This is very much the same analysis as before, except that at at (a+bi)t L(e cos(bt)) + iL(e sin(bt)) = L(e )= Z ∞ e−st e(a+bi)t dt = 0 Z ∞ e −((s−a)−bi)t 0 1 dt = − e−((s−a)−bi)t (s − a) − bi t→∞ 0 As t → ∞, the exponential term will go to zero as long as s − a > 0, or s > a. If that is true, then we have: = 1 b s−a = + i (s − a) − bi (s − a)2 + b2 (s − a)2 + b2 From this, we get: L(eat cos(bt)) = s−a (s − a)2 + b2 4 L(eat sin(bt)) = b (s − a)2 + b2 4. Show that, if f (t) is bounded (that is, there is a constant A so that |f (t)| ≤ A for all t), then f is of exponential order (do this by finding K, a and M from the definition). SOLUTION: If f (t) is bounded, then |f (t)| ≤ A = A · e0·t for all t. 5. If the function is of exponential order, find the K, a, M from the definition. Otherwise, state that it is not of exponential order. Something that may be handy from algebra: A = eln(A) . (a) sin(t) SOLUTION: sin(t) is bounded by 1, so K = 1, a = 0, and M is irrelevant (true for all t). (b) tan(t) SOLUTION: Since the tangent function has vertical asymptotes, tan(t) is not of exponential order. (c) t3 SOLUTION: Consider t > 0: 3 t3 = t3 = eln(t ) = e3 ln t ≤ e3t Therefore, K = 1, a = 3 and M = 0 2 (d) et SOLUTION: Not of exponential order, since we’re raising t to a polynomial power (larger than 1). (e) 5t SOLUTION: t 5t = eln(5 ) = eln(5) t so K = 1, a = ln(5) and M = 0. (f) tt SOLUTION: tt is not of exponential order, since tt = et ln(t) and t ln(t) > at for any constant a. 6. Use complex exponentials to find the Laplace transform of t sin(at). SOLUTION: Using the definition, we’ll consider the imaginary part of the following integral: Z ∞ Z ∞ −st ait e te dt = te−(s−ai)t dt 0 0 5 Using integration by parts, + t e−(s−ai)t − 1 −1/(s − ai) e−(s−ai)t + 0 1/(s − ai)2 e−(s−ai)t ⇒ e −(s−ai)t t 1 − − s − ai (s − ai)2 ! The term in the parentheses will go to zero as long as the exponential goes to zeroWhich it will as long as s > 0. In that case, the integral becomes: 1 1 = 2 2 2 (s − ai) (s − a ) − 2asi When we multiply by the conjugate, the denominator will become: (s2 − a2 )2 + 4a2 s2 = s4 − 2a2 s2 + a4 + 4a2 s2 = s4 + 2a2 s2 + a4 = (s2 + a2 )2 so that finally we get: 1 s 2 − a2 2as = + i (s − a)2 (s2 + a2 )2 (s2 + a2 )2 Therefore, our final answer is the imaginary part of this, 2as + a2 ) (s2 For future reference, you might verify that this expression is actually: a d 2 ds s + a2  (−1)  which is how we will be computing this transform later... 6