Problem #7.11, MA598R, Summer 2009, Weigel
11.) Compute the Fourier transform of $ H(x) = (4\pi)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4}}, x \in {\mathbb R}^n. $
Solution: Following the hint, we consider the integral $ \phi ( \xi ) = \frac{1}{\sqrt{4\pi}}\int_{{\mathbb R}} \cos (2\pi x \xi )e^{-\frac{x^2}{4}}dx. $
Lemma 1: $ \phi $ is differentiable with respect to $ \xi $ (in the sense that we can differentiate inside the integral).
Proof of Lemma 1: (Robert said he'd write this up.)
Form the difference quotient as follows, and note, letting $ f(x,\xi) = e^{-x^2/4} \cos(2\pi\xi x) $, that the only part of this function depending on $ \xi $ is $ \cos(2\pi\xi x) $, and the function has a continuous partial derivative with respect to $ \xi $. So by the Mean Value Theorem, for some $ \eta_{h} \in (0, h) $, $ \int_{R} \frac{\left(f(x+h) - f(x)\right) \cos(2\pi\xi x)}{h} dx = \int_{R} \frac{\partial f}{\partial \xi}(x,\xi + \eta_h) dx $
But, we know that:
$ |e^{-x^2/4} 2\pi x \sin(2\pi\xi x)| \leq e^{-x^2/4} 2\pi |x| $, and the thing on the right is indeed $ L^1 $ hence we choose as our dominating function $ e^{-x^2/4} 2\pi |x| $, hence, by the Lebesgue Dominated Convergence Theorem, we may say:
$ \lim_{h\rightarrow 0^+} \int_{R} \frac{\partial f}{\partial \xi}(x,\xi + \eta_h) dx = \int_{R} \lim_{h\rightarrow 0^+} \frac{\partial f}{\partial \xi}(x,\xi + \eta_h) dx = \int_{R} \frac{\partial f}{\partial \xi}(x,\xi) dx $
And hence we can pass differentiation with respect to $ \xi $ inside the integral.
Lemma 2: $ \phi $ satisfies the differential equation $ \phi '( \xi ) = -8\pi^2\xi \phi ( \xi ). $
Proof of Lemma 2: We have $ \phi ( \xi ) = \frac{1}{\sqrt{4\pi}}\int_{{\mathbb R}} \cos (2\pi x \xi )e^{-\frac{x^2}{4}}dx. $ From Lemma 1, we know $ \phi '( \xi ) = \frac{2\pi}{\sqrt{4\pi}}\int_{{\mathbb R}} \sin (2\pi x \xi )(-x)e^{-\frac{x^2}{4}}dx. $
We seek to evaluate this integral using integration by parts. Take $ u = \sin (2\pi x \xi ) \implies du = 2\pi \xi \cos(2\pi x \xi )dx, $ and $ dv = -x e^{-\frac{x^2}{4}}dx \implies v = 2e^{-\frac{x^2}{4}}. $ Thus, we see that $ \phi '( \xi ) = \frac{2\pi}{\sqrt{4\pi}}\left( 2e^{-\frac{x^2}{4}}\sin (2\pi x \xi )|_{-\infty}^{\infty} - \int_{{\mathbb R}} 2e^{-\frac{x^2}{4}} (2\pi \xi )\cos (2\pi x \xi )dx \right), $ or
$ \phi '( \xi ) = \frac{2\pi}{\sqrt{4\pi}}\left( -4\pi \xi \int_{{\mathbb R}} e^{-\frac{x^2}{4}} \cos (2\pi x \xi )dx \right) = -8\pi^2 \xi \left( \frac{1}{\sqrt{4\pi}}\int_{{\mathbb R}} \cos (2\pi x \xi )e^{-\frac{x^2}{4}}dx \right) $
$ = -8\pi^2\xi \phi ( \xi ), $ as desired. This completes the proof of Lemma 2.
Now, $ \phi '( \xi ) = -8\pi^2\xi \phi ( \xi ), $ so $ \frac{\phi '(\xi )}{\phi (\xi )} = -8\pi^2 \xi. $ Integrating both sides, we see $ \ln (\phi (\xi )) = -4\pi^2 \xi^2, $ or $ \phi (\xi) = e^{-4\pi^2 \xi^2}. $
Now, in a single dimension, we know that:
$ \phi (\xi) = \int_{R} e^{-x^2/4} e^{-2\pi\imath x \xi} dx = e^{-4\pi^2 \xi^2} $
So in n dimensions, we have:
$ \phi (\xi) = \int_{R^n} e^{-|x|^2/4} e^{-2\pi\imath <x, \xi>} dx = \int_{R^n} e^{- \sum_{j = 1}^{n}\left(x_{j}^{2}/4 + 2\pi\imath x_j \xi_j\right)} dx_1 \ldots dx_n $
$ = \int_{R^n} \prod_{j = 1}^{n} e^{- x_{j}^{2}/4 - 2\pi\imath x_j \xi_j} dx_1 \ldots dx_n $
$ = \prod_{j = 1}^n \int_{R} e^{-x_{j}^{2}/4 - 2\pi\imath x_j \xi_j} dx_j $
$ = \prod_{j = 1}^{n} e^{-4\pi^2 \xi_j^2} $, using the result already proven in one dimension. Hence, simplifying a bit,
$ \phi(\xi) = e^{-4\pi^2 |\xi|^2} $
P.S. Since Robert hadn't written up the proof of Lemma 1 by 15 till 10, I decided to pick it up as well. Any errors are blamed on me and my inferior typing/math skills.
-Nick Stull
Rest of it courtesy of Daniel Frederick
