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Our final example uses a slightly different idea. Example 6.9. Show that, if a is real and a> 0 , integraldisplay π 0 a a 2 + sin 2 θ = π (1 + a 2 ) 1 / 2 . There is a mixture of good and bad news about contour integration. (1) Most examples (particularly at 1B level) are based on combining a limited number of tricks. If you are stuck, try to identify parts of the problem which you have met before. (2) The only way, for most people, to become fluent in contour integration is to do lots of examples yourself. (3) Almost every book on complex analysis in your college library 6 will contain a chapter with a large collection of worked examples for you to take as model. 7 Fourier transforms Many systems in nature, engineering and mathematics are linear and allow us to build complex solutions as linear combinations of simpler solutions. Thus, for example, light and sound may be considered as a mixture of simple, single frequency waves. Mathematically we start by considering a single frequency wave e iωt , we then consider a sum of a finite number of such simple waves n summationdisplay j =1 a j e j t with a j C and are then driven to consider the integral analogue integraldisplay −∞ F ( ω ) e iωt dω. To emphasise the connection with Fourier series (see the course Mathe- matical Methods, C10) we use the following definition. 6 Often an architectural gem and well worth visiting for its own sake. 17
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Definition 7.1. If f : R C is reasonably well behaved, we define ˆ f ( λ ) = integraldisplay −∞ f ( t ) e iλt dt, and call the function ˆ f : R C the Fourier transform . This is a methods course, so we shall not go into what is meant by good behaviour. However, the condition that f , f and f ′′ are continuous and t 2 f ( t ) , t 2 f ( t ) , t 2 f ′′ ( t ) 0 as | t | → ∞ are amply sufficient for our purpose (much less is required, but there always has to be some control over behaviour towards infinity). The following results form part of the grammar of Fourier transforms. Lemma 7.2. (i) If a R , let us write f a ( t ) = f ( t a ) . Then ˆ f a ( λ ) = e iaλ ˆ f ( λ ) . (Translation on one side gives phase change on other.) (ii) If K R and K> 0 , let us write f K ( t ) = f ( Kt ) . Then ˆ f K ( λ ) = K 1 ˆ f ( λ/K ) . (Narrowing on one side gives broadening on the other.) (iii) ˆ f ( λ ) = ( f )ˆ( λ ) . (iv) ( ˆ f ) ( λ ) = i ˆ F ( λ ) where F ( t ) = tf ( t ) . (v) ( f )ˆ( λ ) = ˆ f ( λ ) . The next result is both elegant and important. Lemma 7.3. We have integraldisplay −∞ f ( t g ( t ) dt = integraldisplay −∞ ˆ f ( λ ) g ( λ ) dλ. Taking g ( λ ) = exp( ( K 1 λ ) 2 / 2) and allowing K → ∞ , we obtain the key inversion formula. Theorem 7.4 (Inversion formula) . We have f ˆˆ( t ) = 2 πf ( t ) . In other words, f ( t ) = 1 2 π integraldisplay −∞ ˆ f ( ω ) e iωt dω. Thus we can break down any (well behaved) function into its constituent frequencies and then reconstruct it.
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