# Example 3 evaluate z 2 π 1 sin θ 2 dθ answer we

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Example 3 . Evaluate Z 2 π 0 1 sin θ + 2 . Answer . We put z = e and get Z 2 π 0 1 sin θ + 2 = Z C 1 ( z - 1 z ) 1 2 j + 2 dz jz where C is the unit circle. Now Z C dz 2 jz + ( z 2 - 1) 1 2 = Z C 2 dz z 2 - 1 + 4 jz . The roots of the bottom are z = - 4 j ± - 16 + 4 2 = ( - 2 ± 3) j, so z 2 + 4 jz - 1 = ( z + (2 + 3) j )( z + (2 - 3) j ) . C 153
Observe that ( - 2 + 3) j is inside C , but ( - 2 - 3) j is not, since ( | ( - 2 + 3) j | = 2 - 3 < 1 , | ( - 2 - 3) j | = 2 + 3 > 1 . We also note that the pole at ( - 2 + 3) j is simple . Then Z C 2 z 2 + 4 jz - 1 dz = 2 πj 2 2 3 j = 2 π 3 . As a final comment, we recall that f ( x ) is even iff f ( x ) = f ( - x ). Suppose f ( x ) is even, then Z 0 f ( x ) dx = 1 2 Z -∞ f ( x ) dx. We can thus evaluate R 0 f ( x ) dx by evaluating R -∞ f ( x ) dx and dividing by 2. As an example consider R 0 1 1+ x 2 dx . Here f ( x ) = 1 1+ x 2 for f ( x ) = f ( - x ) and f is even . We thus have R 0 1 1+ x 2 dx = 1 2 · π from Example 1. Note also that f ( x ) is odd if f ( x ) = - f ( - x ), so Z -∞ f ( x ) dx = lim R →∞ Z R - R f ( x ) dx but Z R - R f ( x ) dx = Z 0 - R f ( x ) dx + Z R 0 f ( x ) dx = - Z R 0 f ( x ) dx + Z R 0 f ( x ) dx = 0 , and R -∞ f ( x ) dx = 0! Unfortunately, we can’t do R 0 f ( x ) dx if f ( x ) is odd . 154
III.h Inverting the Laplace Transform and Applications We now come one of the main reasons for all the work we have done: inverting the Laplace Transform. The basic idea—constructing suitable paths in the complex plane—is similar to the previous section. You may recall that given y = f ( t ) (usually t is actually time), L ( f )( s )—the Laplace Transform of f —is given by L ( f )( s ) = Z 0 e - st f ( t ) dt. In elementary courses, s is thought of as being a real number, but in practical problems this is not the case and you should think of s as being complex. Indeed, s = jw with w = frequency is of particular significance. So we wish to think as L ( f )( s ) being given for any complex s . The formulas you hopefully recall hold unchanged for s complex. In particular, L ( e at )( s ) = 1 / ( s - a ) if Re ( s ) > Re ( a ). Now let f ( s ) be the transform of some function h ( t ). We know f ( s ) and wish to find h ( t ), i.e., we wish to invert the transform. In practice, f ( s ) will have some poles at some points s 1 , . . ., s m of the complex plane. As is often the case in practice we start by assuming s 1 , . . ., s m all lie in the left half plane. (We shall deal with the other case later.) We construct the path C = C 1 + C 2 with C 2 a semicircle of radius R as shown. C 2 C R 1 s 1 s 2 s m s 3 s j R - j R Think of R as being large. If s is any point inside the semicircle (as shown), then 155
since f is analytic inside and on C we have f ( s ) = 1 2 πj Z C f ( z ) z - s dz. Think of s as being temporarily fixed, and assume | f ( z ) | → 0 as | z | → ∞ . This is the case for most “practical” functions but not for all. The latter case is more complicated and we do not deal with it in this course. Look at C 2 : if z is on C 2 then | f ( z ) | → 0 as | z | (i.e., R ) approaches , and 1 / | z - s | also goes to zero as | z | → ∞ . One can formally show that as a consequence R C 2 f ( z ) / ( z - s ) dz 0 as R → ∞ , and a detailed proof is given in theoretical courses. So, since f ( s ) never changes as R → ∞ , we get f ( s ) = 1 2 πj Z C f ( z ) z - s dz = 1 2 πj Z C 1 f ( z ) z - s dz + Z C 2 f ( z ) z - s dz -→ 1 2 πj Z