3 Fourier and Laplace Transforms The complex exponentials exp(i2⇡nx/L) are orthonormal and easy to dif- ferentiate (and to integrate), but they are periodic with period L. If one wants to represent functions that are not periodic, a better choice is the complex exponentials exp(ikx), where k is an arbitrary real number. These orthonormal functions are the basis of the Fourier transform. The choice of complex k leads to the transforms of Laplace, Mellin, and Bromwich. 3.1 Fourier transforms The interval [ L/2,L/2] is arbitrary in the Fourier series pair (2.46) − 1 ei2⇡nx/L L/2 e i2⇡nx/L f(x)= f and f = f(x) − dx. (3.1) n p n p n= L L/2 L X1 Z− What happens when we stretch this interval without limit, letting L ? !1 We may use the nearest-integer function [y] to convert the coefficients f into a function of a continuous variable fˆ(y) f such that fˆ(y)=f n ⌘ [y] n when y n < 1/2. In terms of this function fˆ(y), the Fourier series (3.1) | − | for the function f(x)is 1 n+1/2 ei2⇡[y]x/L ei2⇡[y]x/L f(x)= fˆ(y) dy = 1 fˆ(y) dy. (3.2) p p n= n 1/2 L L X1 Z − Z1 Since [y] and y di↵er by no more than 1/2, the absolute value of the di↵erence between exp(i⇡[y]x/L) and exp(i⇡yx/L) for fixed x is i2⇡[y]x/L i2⇡yx/L i2⇡([y] y)x/L ⇡ x e e = e − 1 | | (3.3) − − ⇡ L which goes to zero as L . So in this limit, we may replace [y]byy and !1 126 Fourier and Laplace Transforms express f(x) as ei2⇡yx/L f(x)= 1 fˆ(y) dy. (3.4) pL Z1 We let y = Lk/(2⇡)sok =2⇡y/L and find for f(x) the integral Lk eikx L L Lk dk f(x)= 1 fˆ dk = 1 fˆ eikx . (3.5) 2⇡ pL 2⇡ 2⇡ 2⇡ p2⇡ Z1 ✓ ◆ Z1r ✓ ◆ Now in terms of the Fourier transform f˜(k) defined as L Lk f˜(k)= fˆ (3.6) 2⇡ 2⇡ r ✓ ◆ the integral (3.5) for f(x) is the inverse Fourier transform dk f(x)= 1 f˜(k) eikx . (3.7) p2⇡ Z1 To find f˜(k), we use its definition (3.6), the definition (3.1) of fn, our ˜ ˆ ˆ formulas f(k)= Lk/(2⇡)f(Lk/(2⇡)) and f(y)=f[y], and the inequality 2⇡[Lk/2⇡]/L k ⇡/2L to write | − p| L/2 i2⇡[ Lk ] x L/2 ˜ L L e− 2⇡ L ikx dx f(k)= f[ Lk ] = f(x) dx f(x)e− . 2⇡ 2⇡ 2⇡ L/2 pL ⇡ L/2 p2⇡ r r Z− Z− This formula becomes exact in the limit L !1 1 ikx dx f˜(k)= f(x) e− (3.8) p2⇡ Z1 and so we have the Fourier transformations 1 ikx dk 1 ikx dx f(x)= f˜(k) e and f˜(k)= f(x) e− . (3.9) p2⇡ p2⇡ Z1 Z1 The function f˜(k)istheFourier transform of f(x), and f(x)isthein- verse Fourier transform of f˜(k). In these symmetrical relations (3.9), the distinction between a Fourier transform and an inverse Fourier transform is entirely a matter of conven- tion. There is no rule for which sign, ikx or ikx, goes with which transform − or for where to put the 2⇡’s. Thus one often sees 1 ikx 1 ikx dx f(x)= f˜(k) e± dk and f˜(k)= f(x) e⌥ (3.10) 2⇡ Z1 Z1 3.1 Fourier transforms 127 as well as 1 ikx dk 1 ikx f(x)= f˜(k) e± and f˜(k)= f(x) e⌥ dx. (3.11) 2⇡ Z1 Z1 One often needs to relate a function’s Fourier series to its Fourier trans- form. So let’s compare the Fourier series (3.1) for the function f(x) on the interval [ L/2,L/2] with its Fourier transform (3.9) in the limit of large L − setting kn =2⇡n/L =2⇡y/L 1 ei2⇡nx/L 1 eiknx dk f(x)= f = f = 1 f˜(k) eikx . (3.12) n p n p p n= L n= L 2⇡ X1 X1 Z1 ˆ ˜ Since fn = f(y)=f[y], by using the definition (3.6) of f(k), we have Lk 2⇡ f = f = f = fˆ(y)=fˆ = f˜(k). (3.13) n [n] [y] 2⇡ L ✓ ◆ r Thus, to get the Fourier series from the Fourier transform, we multiply the series by 2⇡/L and use the Fourier transform at kn divided by p2⇡ 1 1 2⇡ 1 f˜(k ) f(x)= f eiknx = n eiknx. (3.14) p n L p L n= n= 2⇡ X1 X1 ˜ Going the other way, we set f(k)= L/2⇡fn = L/2⇡f[Lk/2⇡] and find dk pL f[Lk/p2⇡] f(x)= 1 f˜(k) eikx = 1 eikxdk. (3.15) p2⇡ 2⇡ pL Z1 Z1 Example 3.1 (The Fourier Transform of a Gaussian Is a Gaussian). The Fourier transform of the gaussian f(x)=exp( m2 x2)is − 1 dx ikx m2 x2 f˜(k)= e− e− . (3.16) p2⇡ Z1 We complete the square in the exponent: k2/4m2 1 dx m2 (x+ik/2m2)2 f˜(k)=e− e− . (3.17) p2⇡ Z1 As we’ll see in Sec. 5.14 when we study analytic functions, we may replace x by x ik/2m2 without changing the value of this integral. So we can drop − the term ik/2m2 in the exponential and get k2/4m2 1 dx m2 x2 1 k2/4m2 f˜(k)=e− e− = e− . (3.18) p2⇡ p2 m Z1 128 Fourier and Laplace Transforms Thus the Fourier transform of a gaussian is another gaussian 1 dx ikx m2 x2 1 k2/4m2 f˜(k)= e− e− = e− . (3.19) p2⇡ p2 m Z1 But the two gaussians are very di↵erent: if the gaussian f(x)=exp( m2x2) − decreases slowly as x because m is small (or quickly because m is big), !1 then its gaussian Fourier transform f˜(k)=exp( k2/4m2)/mp2 decreases − quickly as k because m is small (or slowly because m is big). !1 Can we invert f˜(k) to get f(x)? The inverse Fourier transform (3.7) says 1 dk ikx 1 dk 1 ikx k2/4m2 f(x)= f˜(k) e = e − . (3.20) p2⇡ p2⇡ mp2 Z1 Z1 By again completing the square in the exponent m2x2 1 dk 1 (k i2m2x)2/4m2 f(x)=e− e− − (3.21) p2⇡ mp2 Z1 and shifting the variable of integration k to k + i2m2x,wefind m2x2 1 dk 1 k2/(4m2) m2x2 f(x)=e− e− = e− (3.22) p2⇡ mp2 Z1 which is reassuring. Using (3.18) for f˜(k) and the connections (3.12–3.15) between Fourier series and transforms, we see that a Fourier series for this gaussian is in the limit of L x 2 2 1 2 2 m x 2⇡ 1 k /(4m ) iknx f(x)=e− = e− n e (3.23) L p n= 4⇡m X1 in which kn =2⇡n/L. 3.2 Fourier transforms of real functions If a function f(x) is real, then the complex conjugate of its Fourier transform (3.8) 1 dx ikx f˜(k)= f(x) e− (3.24) p2⇡ Z1 is its Fourier transform evaluated at k − 1 dx ikx f˜⇤(k)= f(x) e = f˜( k). (3.25) p2⇡ − Z1 3.2 Fourier transforms of real functions 129 It follows (exercise 3.1) that a real function f(x) satisfies the relation 1 f(x)= 1dk 1 f(y) cos k(y x) dy. (3.26) ⇡ 0 − Z Z1 If f(x) is both real and even, then 2 f(x)= 1cos kx dk 1f(y) cos ky dy (3.27) ⇡ Z0 Z0 if it is even, and 2 f(x)= 1sin kx dk 1f(y)sinky dy (3.28) ⇡ Z0 Z0 if it is odd (exercise 3.2). Example 3.2 (Dirichlet’s Discontinuous Factor). Using (3.27), one may write the square wave 1 x < 1 | | f(x)= 1 x =1 (3.29) 8 2 | | 0 x > 1 < | | as Dirichlet’s discontinuous factor: 2 sin k cos kx f(x)= 1 dk (3.30) ⇡ k Z0 (exercise 3.3). Example 3.3 (Even and Odd Exponentials). By using the Fourier-transform formulas (3.27 & 3.28), one may show that the Fourier transform of the even exponential exp( β x )is − | | β x 2 1 β cos kx e− | | = dk (3.31) ⇡ β2 + k2 Z0 while that of the odd exponential x exp( β x )/ x is − | | | | x β x 2 1 k sin kx e− | | = dk (3.32) x ⇡ β2 + k2 | | Z0 (exercise 3.4). 130 Fourier and Laplace Transforms 3.3 Dirac, Parseval, and Poisson Combining the basic equations (3.9) that define the Fourier transform, we may do something apparently useless: we may write the function f(x)in terms of itself as 1 dk ikx 1 dk ikx 1 dy iky f(x)= f˜(k) e = e e− f(y). (3.33) p2⇡ p2⇡ p2⇡ Z1 Z1 Z1 Let’s compare this equation dk f(x)= 1 dy 1 exp[ik(x y)] f(y) (3.34) 2⇡ − Z1 ✓Z1 ◆ with one (2.116) that describes Dirac’s delta function f(x)= 1 dy δ(x y) f(y).