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I have to calculate the Fourier transform of the following distribution: $$\langle F_a\mid\phi\rangle = \int_{-\infty}^\infty \cos(ax) \phi(x) \, dx, \phi \in \mathcal S (\mathbb{R})$$ Would it be wrong to calculate instead the Fourier transform on: $$F_a=\cos(ax) \text{?}$$

In both cases, can you show me a complete solution?

Michael Hardy
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agneau
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1 Answers1

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The fact is that $F_{a}\notin L^{1}({\bf{R}})$, so the integral $\displaystyle\int_{\bf{R}}\cos(ax)e^{-2\pi i\xi\cdot x}dx$ does not exist. Besides that, $F_{a}\notin L^{2}({\bf{R}})$, so the general Fourier transform also does not exist. So we can only speak about the distributional Fourier transform.

For the calculation, note that $\cos(ax)=\dfrac{1}{2}(e^{iax}+e^{-iax})$ and hence for a test function $\varphi\in S({\bf{R}})$, we have \begin{align*} \left<\widehat{\cos(ax)},\varphi\right>&=\left<\cos(ax),\widehat{\varphi}\right>\\ &=\dfrac{1}{2}\left<e^{iax},\widehat{\varphi}\right>+\dfrac{1}{2}\left<e^{-iax},\widehat{\varphi}\right>\\ &=\dfrac{1}{2}\int_{\bf{R}}e^{iax}\widehat{\varphi}(x)dx+\dfrac{1}{2}\int_{\bf{R}}e^{-iax}\widehat{\varphi}(x)dx\\ &=\dfrac{1}{2}(\widehat{\varphi})^{\vee}(a/2\pi)+\dfrac{1}{2}(\widehat{\varphi})^{\vee}(-a/2\pi)\\ &=\dfrac{1}{2}(\varphi(a/2\pi)+\varphi(-a/2\pi))\\ &=\dfrac{1}{2}\left<\delta_{a/2\pi},\varphi\right>+\left<\delta_{-a/2\pi},\varphi\right>\\ &=\left<\dfrac{1}{2}(\delta_{a/2\pi}+\delta_{-a/2\pi}),\varphi\right>, \end{align*} so \begin{align*} \widehat{\cos(ax)}=\dfrac{1}{2}(\delta_{a/2\pi}+\delta_{-a/2\pi}) \end{align*} in the distributional sense.

user284331
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  • Ok I see. May you show me how to calculate the distributional Fourier transform? – agneau Feb 15 '18 at 13:41
  • I have added the calculation. – user284331 Feb 15 '18 at 18:42
  • thank you, you are very helpful. May you take a look also at this question? https://math.stackexchange.com/questions/2650556/limit-and-derivative-of-f-n-f-n-phi-int-0-infty-dx-sin2nx-phix – agneau Feb 15 '18 at 18:52
  • I've never seen before the "v" symbol with which you "elevated" the $\phi$ function. What does it stands for? – agneau Feb 17 '18 at 10:55
  • is it the antitransform symbol? – agneau Feb 17 '18 at 11:43
  • I understand that you used the inversion theorem but I can't understand why you divided also for $2 \pi$ – agneau Feb 17 '18 at 12:02
  • Because the formula reads like this: $\widehat{f}(\xi)=\displaystyle\int f(x)e^{-2\pi ix\cdot\xi}dx$ and $(\widehat{f})^{\vee}=f$ for sufficiently good functions. – user284331 Feb 17 '18 at 18:05
  • Perfect. I was confused by the fact that in my book is used a different convention for the fourier transform ($\frac{e^{-ixk}}{\sqrt{2 \pi}}$ – agneau Feb 17 '18 at 18:53
  • writing the transform like this I got: $\frac{1}{2\sqrt{2\pi}}(<\delta_a,\phi>+<\delta_{-a}, \phi>$ – agneau Feb 17 '18 at 18:55
  • Really? But I think the inversion formula from your book should sound a little bit different and the final answer for the question should be the same. – user284331 Feb 17 '18 at 19:02
  • I'll write you the definitions of transform and antitrasform: $$(\mathcal F \phi)(k)=\int_{\mathbb{R}} \frac{dx}{\sqrt{2\pi}} e^{-ikx} \phi(x)$$ $$(\mathcal F^{-1} \phi)(k)=\int_{\mathbb{R}} \frac{dx}{\sqrt{2\pi}} e^{ikx} \phi(x)$$ – agneau Feb 17 '18 at 19:40
  • And the inversion theorem states: $$\phi(x)= \int_{\mathbb{R} } \frac{dk}{\sqrt{2 \pi}} e^{ikx} (\mathcal F \phi)(k)$$ – agneau Feb 17 '18 at 19:44
  • I guess that your definition is related to the physical meaning of fourier transform as a function that brings a function from the time domain to the frequency domain. And as for the frequency there are these relations that may have something to do: $$\mathcal f=\frac{1}{T}=\frac{2\pi}{\omega}$$ – agneau Feb 17 '18 at 19:48
  • Ah... it may depend on how do you define the Fourier transform for distributions, the answer may vary for some constant, but the Dirac Delta must be there, this is the crucial one. – user284331 Feb 17 '18 at 19:49
  • but I don't know, the Fourier transform is a total mystery to me. If you have some suggestions/links/articles for me to read you'd be welcome. I really want to figure out something but I'm struggling – agneau Feb 17 '18 at 19:50
  • I was reading Loukas Grafakos, Classicla Fourier Analysis, but this is a heavy favour for analysis people, no any physics stuff inside, not sure if this is a book that you will like. – user284331 Feb 17 '18 at 19:50
  • Yours it's an intimidating book... I'll stick to my physics book! – agneau Feb 17 '18 at 19:54
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    Sorry, I know nothing about physics so I have no idea for which physics favour book is the right one. – user284331 Feb 17 '18 at 20:02