PropertiesoftheClassicalFourier Transform; Some Examples De nition oftheClassicalFourier Transform Let f x be a possibly complex-valued function de ned for ,1 x and square integrable, i.e., jf x j2 dx ,1 Z 1: There are certain regularity requirements inherent in this de nition which we not want to get into right now; we comment further on this in the section on proof ofFourier transform propertiesThe transform produces from f x , another function, f^ , by the formula Z 1 ^ p f = e,ix f x dx: 2 ,1 It turns out that the transform is invertible, with the inverse transform being given by Z 1 eix f^ d: f x = p 2 ,1 Minor variations common in the literature involve replacement ofthe factor p common to both of our formulas with in one ofthe formulas and just in the other formula The factor p plays much the same role for the classical, or continuous Fourier transform as the factor pN does in the case ofthe discrete Fourier transform With the transform as we have de ned it, the function f^ is square integrable just in case f x is square integrable and we have the Plancherel identity 2 2 j f x j dx ,1 Z = 1 f^ d: ,1 Z It is also true that if f x and g x are both square integrable functions and f^ and g^ are their respective transforms, then hf; gi f x g x dx ,1 Z = Z f^ ^ g d ,1 = f^; g^ D E : The second property reduces to the rst when we set g x = f x , of course We express both of these properties by saying that theFourier transform is unitary Fourier Transform PropertiesThepropertiesoftheFourier transform are much the same as, but not identical to, thepropertiesofthe Laplace transform This is to be expected because, as shown in the section on the relationship between the two, the Laplace transform is just a special case oftheFourier transform together with a change of variable in the complex plane In many cases the proofs oftheFourier transform properties are essentially the same as those given for the Laplace transform; we not repeat those here The property of linearity is more or less obvious; we refer the reader to the Laplace transform discussion In the subsequent discussion oftheFourier transform and its properties we will use the symbols f^ or calF f , whichever is more convenient under the circumstances Property I: Fourier Transform of eiax f x We have iax calF e = Z 1 f x = p e,ix eiax f x dx 2 ,1 ,i ,a x e f x dx ,1 Z = calF f Property II: Fourier Transform of ix n f x , a : Z 1 calF ix f x = p e,ix ix n f x dx 2 ,1 n = ,1 n dn d n ! Z n , ix p ,1 e f x dx = ,1 n dd n calF f : 2 Property III: Fourier Transform of F x = 0x f y dy If f x and lies in L ,1; 1 and, F x , as indicated, lie in L ,1; 1 then R calF f 2 using integration by parts, we have Z e,ix Z x f y dy dx calF F = p 2 ,1 ! Z e,ix calF f : f x dx = = p i 2 ,1 i Property IV: Fourier Transform of f x ; f k x If f x and f x lie in L ,1; 1 then calF f lies in L ,1; 1 and, again using integration by parts, calF f Z 1 = p e,ix f x dx 2 ,1 Z 1 i e,ix f x dx = i calF f : =p 2 ,1 This process can be repeated to see that if f x ; f x ; :::; f k x all lie in L ,1; 1 then j calF f lies in L ,1; 1 ; j = 1; 2; :::; k and calF f j = i j calF f ; j = 1; 2; :::; k: Property V: Fourier Transform of f x , a The behavior oftheFourier transform with respect to shifted functions is simpler than the corresponding behavior ofthe Laplace transform No distinction needs to be made between the de nitions of right and left hand shifts; we simply observe that for any real number a Z 1 calF f x , a = p e,ix f x , a dx 2 ,1 Z = e,ia p1 e,i x,a f x , a d x , a = e,ia calF f : 2 ,1 Property VI: Fourier Transform ofthe Convolution f g x The convolution product of two functions f x ; g x in L ,1; 1 is de ned by f g x = f y g x , y dy: ,1 Z The integral is de ned for all real x because f y and g x,y lie in L ,1; 1 if f x and g x lie in that space and the Schwarz inequality then shows the product f y g x , y to be absolutely integrable However, f g x does not necessarily lie in L ,1; 1 In fact we have the result: f g x L ,1; 1 if and only if the product calF f calF g L ,1; 1 and, with the change of variable r = x , y, Z e,ix Z f y g x , y dy dx calF f g x = p ,1 2 ,1 Z Z 1 e,i r y f y g r dy dr =p 2 ,1 ,1 ! ! Z p Z ,iy p ,1 e,ir f r dr = 2 p e f y dy 2 ,1 2 p = 2 calF f calF g : 2 2 + Computing theFourier Transform A major distinction between theFourier transform and its Laplace counterpart lies in the fact that many, in some sense most, ofthe familiar elementary functions not have Fourier transforms in the standard sense of that term That is true because, in most cases, they not decay rapidly enough at 1 for theFourier integral to be de ned Thus we not have standard Fourier transforms, e.g., for f x = xn ; f x = eax ; f x = sin ax, etc it is possible to interpret the transforms ofthe rst and second of these as distributions, however Even the transforms of such functions such as f x = x2 , which are de ned in the standard sense, turn out to involve combinations of di erent elementary functions As a consequence, while the Laplace and Fourier transforms are, mathematically, very closely related they are, in fact, used in rather di erent contexts Nevertheless, we begin with some examples where theFourier transform can be computed without too much di culty 1+ Example We compute theFourier transform of f x = e, ax2 In this case Z Z 1 , ix ,ax2 ^ f = p e e dx = p e, ax 2 ,1 2 ,1 +ix ; a dx Z Z 2 ,a x2 i a x, 4a22 , 42a 1 , a p p e e e dx = e,a x i 2a dx: = ,1 2 ,1 2 Using the methods of contour integration in the complex plane one can show that Z Z 2 e,a x i 2a dx = e,ax dx ,1 ,1 for any real To compute the last integral we note that + + + Z Z 1 ,ax2 2 e dx = e,ax dx e,ay dy ,1 ,1 ,1 Z Z Z 2 Z , a x2 +y e dx dy = e,ar r dr d ; ,1 ,1 0 Z = the last identity following from conversion to polar coordinates in the plane Then Z Z Z Z 1 , ar e r dr d = e,ar 2ar dr d 2a Z = a , drd e,ar2 dr = a ,e,ar2 = a : 2 0 0 Thus we obtain r , 2 2 ^ f = p e 4a = p e, 4a : 2 a 2a When we take a = corresponding to f x = e, x2 we nd, signi cantly, 2 that f^ = e, 2 This shows that f x = e, x2 is an eigenfunction oftheFourier transform operator corresponding to the eigenvalue = 1 Example Let us de ne f x Then = e,ax ; x eax ; x 0; 0: Z Z e,ix e,ax dx = p1 e,ix eax dx + p 2 ,1 2 Z Z 1 e a,i x dx + p e, a i x dx =p 2 ,1 2 s ! = p1 a ,1 i + a +1 i = 2 a +a : 2 f^ 0 + 2 Example Observing that the result ofthe previous example is unchanged if we replace , by xi, we have, with f x as given in that example, s a = p1 2 ix e f x dx: ,1 Z + Interchanging the roles of and x we have a Z r f d: p g x = a +x 2 ,1 Applying theFourier transform to both sides and using property i stated in the introductory Fourier transform section, we obtain r Z f = p e,ix g x dx: 2 ,1 a2 2 Thus, with g^ we have r 8q f = : e,ax ; x q eax ; x Z 1 a g^ = p e,ix a +x 2 ,1 2 0; ; dx: This example illustrates the point, made earlier, that theFourier transform of a function having a simple algebraic expression may turn out not to be expressible in terms of a single elementary function In other cases theFourier transform may have no expression in terms of any number of elementary functions ... both of these properties by saying that the Fourier transform is unitary Fourier Transform Properties The properties of the Fourier transform are much the same as, but not identical to, the properties. .. Fourier transform together with a change of variable in the complex plane In many cases the proofs of the Fourier transform properties are essentially the same as those given for the Laplace transform;... properties of the Laplace transform This is to be expected because, as shown in the section on the relationship between the two, the Laplace transform is just a special case of the Fourier transform