(New page: 11.) Compute the Fourier transform of <math>H(x) = (4\pi)^{-\frac{n}{2}}e^{-\frac{|x|^2}{4}}, x \in {\mathbb R}^n.</math><br> <i>Solution:</i> Following the hint, we consider the integral ...) |
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<b>Lemma 1:</b> <math>\phi</math> is differentiable with respect to <math>\xi</math> (in the sense that we can differentiate inside the integral).<br> | <b>Lemma 1:</b> <math>\phi</math> is differentiable with respect to <math>\xi</math> (in the sense that we can differentiate inside the integral).<br> | ||
<i>Proof of Lemma 1:</i> (Robert said he'd write this up.)<br> | <i>Proof of Lemma 1:</i> (Robert said he'd write this up.)<br> | ||
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| + | Form the difference quotient as follows, and note, letting <math>f(x,\xi) = e^{-x^2/4} \cos(2\pi\xi x)</math>, that the only part of this function depending on <math>\xi</math> is <math>\cos(2\pi\xi x)</math>, and the function has a continuous partial derivative with respect to <math>\xi</math>. So by the Mean Value Theorem, for some <math>\eta_{h} \in (0, h)</math>, | ||
| + | <math>\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</math> | ||
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| + | But, we know that: | ||
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| + | <math>|e^{-x^2/4} 2\pi x \sin(2\pi\xi x)| \leq e^{-x^2/4} 2\pi |x|</math>, and the thing on the right is indeed <math>L^1</math> hence we choose as our dominating function <math>e^{-x^2/4} 2\pi |x|</math>, hence, by the Lebesgue Dominated Convergence Theorem, we may say: | ||
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| + | <math>\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</math> | ||
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| + | And hence we can pass differentiation with respect to <math>\xi</math> inside the integral. | ||
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<b>Lemma 2:</b> <math>\phi</math> satisfies the differential equation <math>\phi '( \xi ) = -8\pi^2\xi \phi ( \xi ).</math><br> | <b>Lemma 2:</b> <math>\phi</math> satisfies the differential equation <math>\phi '( \xi ) = -8\pi^2\xi \phi ( \xi ).</math><br> | ||
<i>Proof of Lemma 2:</i> We have <math>\phi ( \xi ) = \frac{1}{\sqrt{4\pi}}\int_{{\mathbb R}} \cos (2\pi x \xi )e^{-\frac{x^2}{4}}dx.</math> From Lemma 1, we know <math>\phi '( \xi ) = \frac{2\pi}{\sqrt{4\pi}}\int_{{\mathbb R}} \sin (2\pi x \xi )(-x)e^{-\frac{x^2}{4}}dx.</math> | <i>Proof of Lemma 2:</i> We have <math>\phi ( \xi ) = \frac{1}{\sqrt{4\pi}}\int_{{\mathbb R}} \cos (2\pi x \xi )e^{-\frac{x^2}{4}}dx.</math> From Lemma 1, we know <math>\phi '( \xi ) = \frac{2\pi}{\sqrt{4\pi}}\int_{{\mathbb R}} \sin (2\pi x \xi )(-x)e^{-\frac{x^2}{4}}dx.</math> | ||
We seek to evaluate this integral using integration by parts. Take <math>u = \sin (2\pi x \xi ) \implies du = 2\pi \xi \cos(2\pi x \xi )dx,</math> and <math>dv = -x e^{-\frac{x^2}{4}}dx \implies v = 2e^{-\frac{x^2}{4}}.</math> Thus, we see that <math>\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),</math> or | We seek to evaluate this integral using integration by parts. Take <math>u = \sin (2\pi x \xi ) \implies du = 2\pi \xi \cos(2\pi x \xi )dx,</math> and <math>dv = -x e^{-\frac{x^2}{4}}dx \implies v = 2e^{-\frac{x^2}{4}}.</math> Thus, we see that <math>\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),</math> or | ||
| − | <math>\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 ),</math> as desired. This completes the proof of Lemma 2.<br> | + | <math>\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)</math> |
| − | Now, <math>\phi '( \xi ) = -8\pi^2\xi \phi ( \xi ),</math> so <math>\frac{\phi '(\xi )}{\phi (\xi )} = -8\pi^2 \xi.</math> Integrating both sides, we see <math>\ln (\phi (\xi )) = -4\pi^2 \xi^2,</math> or <math>\phi (\xi) = e^{-4\pi^2 \xi^2}.</math> ( | + | |
| + | <math> = -8\pi^2\xi \phi ( \xi ),</math> as desired. This completes the proof of Lemma 2.<br> | ||
| + | Now, <math>\phi '( \xi ) = -8\pi^2\xi \phi ( \xi ),</math> so <math>\frac{\phi '(\xi )}{\phi (\xi )} = -8\pi^2 \xi.</math> Integrating both sides, we see <math>\ln (\phi (\xi )) = -4\pi^2 \xi^2,</math> or <math>\phi (\xi) = e^{-4\pi^2 \xi^2}.</math> | ||
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| + | Now, in a single dimension, we know that: | ||
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| + | <math>\phi (\xi) = \int_{R} e^{-x^2/4} e^{-2\pi\imath x \xi} dx = e^{-4\pi^2 \xi^2}</math> | ||
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| + | So in n dimensions, we have: | ||
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| + | <math>\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 </math> | ||
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| + | <math>= \int_{R^n} \prod_{j = 1}^{n} e^{- x_{j}^{2}/4 - 2\pi\imath x_j \xi_j} dx_1 \ldots dx_n </math> | ||
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| + | <math>= \prod_{j = 1}^n \int_{R} e^{-x_{j}^{2}/4 - 2\pi\imath x_j \xi_j} dx_j</math> | ||
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| + | <math> = \prod_{j = 1}^{n} e^{-4\pi^2 \xi_j^2}</math>, using the result already proven in one dimension. Hence, simplifying a bit, | ||
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| + | <math>\phi(\xi) = e^{-4\pi^2 |\xi|^2}</math> | ||
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| + | 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. | ||
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| + | -Nick Stull | ||
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| + | Rest of it courtesy of Daniel Frederick | ||
Revision as of 16:58, 29 July 2009
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
