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Singular Fourier Transforms and the Integral Representation of the Dirac Delta Function

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Singular Fourier Transforms and the Integral Representation of the Dirac Delta Function Physics 116C Singular Fourier transforms and the Integral Representation of the Dirac Delta Function Peter Young (Dated: November 10, 2013) I. INTRODUCTION You will recall that Fourier transform, g(k), of a function f(x) is defined by ∞ g(k)= f(x)eikx dx, (1) Z−∞ and that there is a very similar relation, the inverse Fourier transform,1 transforming g(k) back to f(x): 1 ∞ f(x)= g(k)e−ikx dk, (2) 2π Z−∞ One can show that, for the Fourier transform to converge as the limits of integration tend to , we must have f(x) 0 as x . In addition, let’s consider the g (k), the Fourier ±∞ → | | → ∞ 1 ′ transform of the derivative f (x), and call this g1(k), i.e. ∞ ′ ikx g1(k)= f (x)e dx, (3) Z−∞ Integrating by parts gives ∞ ∞ ikx ikx g1(k)= f(x)e (ik) f(x)e dx, (4) −∞ − −∞ h i Z which shows that the standard relation between g1(k) and g(k), g (k)= ikg(k) , (5) 1 − also only holds if f(x) vanishes for x . Hence standard Fourier transforms only apply to | | → ∞ functions which vanish at infinity. 1 Sometimes the Fourier transform and its inverse are made equivalent by incorporating at factor of 1/√2π into the definition of g(k) namely ∞ 1 ikx g(k) = f(x)e dx, √2π Z−∞ ∞ 1 −ikx f(x) = g(k)e dk. √2π Z−∞ 2 Nonetheless, Fourier transforms are so useful that it is desirable to apply them to some functions which do not satisfy this condition. These transforms are known as “singular Fourier transforms” and will need some form of “regularization” to make the integrals converge. NOTE: for this course, the important sections are I, II, III, and V. II. A SINGULAR FOURIER TRANSFORM INVOLVING A DELTA FUNCTION As an example consider f(x) = 1. In order that the Fourier transform g(k) exists, we regularize the integral by putting in the “convergence factor” e−ǫ|x| where ǫ is small and positive. Eventually we will let ǫ tend to zero. Hence we determine the Fourier transform of −ǫ|x| fǫ(x)= e , (6) which is ∞ ikx −ǫ|x| gǫ(k)= e e dx. Z−∞ We separate the integral into the negative-x region and the positive-x region to find 0 ∞ ikx ǫx ikx −ǫx 1 1 2ǫ gǫ(k)= e e dx + e e dx = + = . (7) ik + ǫ ik + ǫ ǫ2 + k2 Z−∞ Z0 − For ǫ 0 gǫ(k), becomes a narrow high peak, the area under which is → ∞ 2ǫ ∞ dk = 2 tan−1(k/ǫ) = 2π. ǫ2 + k2 −∞ Z−∞ It is therefore convenient to define a quantity δǫ(k) by 1 ǫ δǫ(k)= , (8) π ǫ2 + k2 which has unit area under it. + We now consider the limit ǫ 0 , for which δǫ(k) is a representation of what is known as → the Dirac delta function δ(k). This is an “infinitely high, infinitely narrow” peak with unit area under it. It is defined by the two relations δ(x) = 0, (x = 0), (9) 6 δ(x) dx = 1, (if region of integration includes x = 0). (10) Z 3 From these, it is straightforward to prove the following results: δ(x a)f(x) dx = f(a), (11) − Z δ(x) δ(cx) = , (12) c | | where the region of integration in Eq. (11) includes x = a. You should have seen Eqs. (11) and (12) before. If you are unfamiliar with them, you should take the trouble to derive them. From Eqs. (7) and (8) we have ∞ ikx −ǫ|x| e e dx = 2πδǫ(k). (13) Z−∞ This is a Fourier transform, for which I use the following notation: −ǫ|x| FT e 2πδǫ(k) . (14) −→ The integral in Eq. (13) is well defined because the e−ǫ|x| factor ensures convergence. Equations like Eq. (13) are generally used in situations when they are multiplied on both sides by a smooth function of k, u(k) say, and integrated, i.e. ∞ ∞ ikx −ǫ|x| u(k) e e dx dk = 2π u(k)δǫ(k) dk. (15) Z−∞ Z−∞ Z It turns out that this equation is well behaved if ǫ is set to zero on the LHS (and the limit ǫ 0 → is taken on the RHS). This gives ∞ ∞ ∞ u(k) eikx dx dk = 2π u(k)δ(k) dk Z−∞ Z−∞ Z−∞ = 2πu(0) , (16) (which is known as Fourier’s integral). We used Eq. (11) to get the last expression. One is often tempted to set ǫ to zero also in Eq. (13) (i.e. without multiplying it by a smooth function and integrating), in which case we write ∞ eikx dx = 2πδ(k) . (17) Z−∞ However, as it stands, Eq. (17) does not make sense because the integral does not exist. We therefore have to understand Eq. (17) in one of the following two senses: As a stand-alone equation, in which case it has to be regularized by the convergence factor • e−ǫ|x|, so Eq. (17) really means Eq. (13) for ǫ tending to zero (but not strictly zero). 4 Multiplied by a smooth function u(k) and integrated over k as in Eq. (15), in which case • the convergence factor is unnecessary. Equation (17) is then really a shorthand for Eq. (16). This is normally the sense in which we understand Eq. (17). Since Eq. (17) is a Fourier transform, we can write it as 1 FT 2πδ(k) , (18) −→ which should be compared with Eq. (14). The inverse transform of the delta function then gives 1 ∞ f(x)= 2πδ(k)e−ikx dk = 1 , 2π Z−∞ as required. III. APPLICATIONS OF THE INTEGRAL REPRESENTATION OF THE DELTA FUNCTION In this section we give some applications of the integral representation of the delta function, Eq. (17). A. Convolution Theorem In a previous class, you have already met the convolution theorem, that is, if ∞ F (x)= f(y)f(x y) dy, (19) − Z−∞ which is the convolution of f with itself, then G(k), the Fourier transform of F (x), is simply related to g(k), the Fourier transform of f(x), by G(k)= g(k)2 . (20) We will now give a simple alternative derivation of this result using the integral representation of the delta function, and then use this method to obtain a generalized convolution theorem. We can write Eq. (19) as ∞ ∞ F (x)= dx dx f(x )f(x )δ(x + x x). (21) 1 2 1 2 1 2 − Z−∞ Z−∞ Using Eq. (17) we have 1 ∞ ∞ ∞ F (x)= dx dx dkf(x )f(x )eik(x1+x2−x). (22) 2π 1 2 1 2 Z−∞ Z−∞ Z−∞ 5 The integrals over x1 and x2 are now independent of each other and can be carried out, with the result that 1 ∞ ∞ 2 1 ∞ F (x)= f(t)eikt dt e−ikx dk, = g(k)2e−ikx dk, (23) 2π 2π Z−∞ Z−∞ Z−∞ which shows that F (x) is the inverse transform of g(k)2, i.e. that g(k)2 is the Fourier transform of F (x). Hence we have obtained Eq. (20). The derivation immediately generalizes to the case of two different functions f1(x) and f2(x), with Fourier transforms g1(k) and g2(k), namely, the Fourier transform of the convolution ∞ F (x)= f (y) f (x y) dy, 1 2 − Z−∞ is given by G(k) where G(k)= g1(k)g2(k) . (24) We can now generalize this result to the case where the convolution, rather than involving two variables as in Eq. (21), involves n variables, x ,x , ,xn, but with the same constraint that the 1 2 ··· sum must equal some prescribed value x, i.e. ∞ ∞ F (x)= dx dxn f(x ) f(xn) δ(x + + xn x). (25) 1 ··· 1 ··· 1 ··· − Z−∞ Z−∞ Using the integral representation of the delta function, as before, the integrals over the xi decouple, and we find 1 ∞ ∞ n 1 ∞ F (x)= f(t)eikt dt e−ikx dk, = g(k)ne−ikx dk, (26) 2π 2π Z−∞ Z−∞ Z−∞ which shows that the Fourier transform of F (x) is G(k)= g(k)n, (27) a remarkably simple result. Equation (27) is the desired generalization of the convolution theorem, Eq. (20), to n variables. We shall use this result in class to solve a problem in statistics. Note: With the alternative definition of Fourier transforms given in footnote 1, which puts in factors of √2π to make the transform and the inverse transform symmetric with respect to each other, the convolution theorem for n variables is √2πG(k)= √2πg(k)n . (28) h i 6 B. Parseval’s Theorem Related to the convolution theorem is another useful theorem associated with the name of Parseval. (You may recall that there is a Parseval’s theorem for Fourier series, which is actually closely related.) Using the Fourier transform ∞ g(k)= f(x)eikx dx (29) Z−∞ we have ∞ ∞ ∞ ∞ ⋆ ⋆ ik(x1−x2) g(k)g (k) dk = dx1 dx2 dkf(x1)f (x2)e . (30) Z−∞ Z−∞ Z−∞ Z−∞ Doing the integral over k using Eq. (17) gives 2πδ(x x ), so 1 − 2 ∞ ∞ ∞ g(k) 2 dk = 2π dx dx f(x )f ⋆(x )δ(x x ) | | 1 2 1 2 1 − 2 Z−∞ Z−∞ Z−∞ ∞ = 2π f(x) 2 dx, (31) | | Z−∞ which is Parseval’s theorem. Note: With the alternative definition of Fourier transforms, the factor of 2π in Eq. (31) is missing, so there is complete symmetry between the two sides. IV. EXAMPLES OF SINGULAR FOURIER TRANSFORMS INVOLVING A STEP FUNCTION It is also interesting to consider singular Fourier transforms of functions involving the (Heaviside) step function θ(x) 1 x 0 7 0, (x< 0) θ(x)= , 1, (x> 0) which is denoted H(x) in the book.
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