Statistical Physics

B.3 The Fourier Transform of a Gaussian Distribution Function 189Here n is an integer. In particular, the Gaussian integrals for n =0andn =1are∫ ∞√ πdx e −σx2 =(B.11)σ−∞and∫ ∞√ πdxx 2 e −σx2 =−∞2σ .(B.12)3/2Equation (B.11) can be proved in the following way. First we consider thefollowing double integral:∫ ∞−∞∫ ∞dx dy e −σ(x2 +y 2) .−∞(B.13)The integrals with respect to x and y are independent, and so we can writethis equation as ∫ ∞ ∫ ∞dx e −σx2 dy e −σy2 .(B.14)−∞This means that this double integral is equal to the square of the left-handside of (B.11). On the other hand, we can consider the double integral asa surface integral in the xy plane. Using polar coordinates (r, θ) and notingthat x 2 + y 2 = r 2 and dx dy = r dr dθ, we can rewrite the double integral inthe following form:∫ ∞ ∫ 2πdr dθre −σr2 .(B.15)00This integral is easily evaluated, and the result is π/σ. Thus(B.11)hasbeenproved. Equation (B.10) can be derived from (B.11) by differentiating bothsides n times with respect to σ.−∞B.3 The Fourier Transformof a Gaussian Distribution FunctionThe Fourier transform of a Gaussian distribution function also has the formof a Gaussian distribution function. To show this, we consider the followingintegral:∫ ∞−∞dx e kx e −σx2 ==∫ ∞−∞∫ ∞−∞[dx exp −σdx exp(x − k2σ) ]2+ k24σ] √ [−σx 2 + k2 π= /4σ4σ σ ek2 . (B.16)