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.)
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}. $ (Nick said he'd finish up from here.)
