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I want to solve $\mathscr{F}\{e^{ix}\}(\xi)$ in terms of distributions.

This is what I did

$$\mathscr{F}\big(e^{ix}\big)(\xi) = \int e^{ix} e^{i(\xi,x)}\text{d}x$$

But how do I get further from here?. I used the formula in Vladimirov, on p 108, but this seems to me too general as a solution.

Any help appreciated

Luthier415Hz
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    Where did $\phi$ come from? It's not in the left-most part. – md2perpe Mar 05 '24 at 17:05
  • That is the formula . It is a distribution, so $\phi$ is a rapidly decreasing test-function. – Luthier415Hz Mar 05 '24 at 18:37
  • It should also be in the first part, the one that now is $\mathscr{F}\big(e^{ix}\big)(\xi).$ You are confusing the definition of ordinary functions with the one for distributions. The first part should be $\langle \mathscr{F}\big(e^{ix}\big), \phi \rangle.$ – md2perpe Mar 05 '24 at 19:16
  • but this is taken from the book – Luthier415Hz Mar 05 '24 at 19:22
  • Really? If so, it's hopefully a mistake by the author and noted in the errata. – md2perpe Mar 05 '24 at 20:14
  • @md2perpe It is written on p 108, at the first equation of section 6.2 $$\mathscr{F}f(x)=\int f(x)e^{i(\xi,x)}dx$$ . If you insert $e^{ix}$ you get what I wrote. Note the comma in the exponent of the Fourier transform function. – Luthier415Hz Mar 06 '24 at 08:51
  • That formula is correct. It's the definition of the Fourier transform of an $L^1$ function $f.$ But in your post you don't have that right hand side; instead you have an extra integral involving $\phi.$ – md2perpe Mar 06 '24 at 21:06
  • Hei Per, I edited the question, so it is more in accord with your comment. – Luthier415Hz Mar 07 '24 at 08:25

2 Answers2

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Let $f(x)=e^{ix}$ and let $\phi\in \mathbb{S}$. Appealing to the Fourier Inversion Theorem, we have $$\begin{align} \langle \mathscr{F}\{f\},\phi\rangle&=\langle f,\mathscr{F}\{\phi\},\rangle\\\\ &=\int_{-\infty}^\infty e^{ix}\int_{-\infty}^\infty \phi(\xi)e^{i\xi x}\,d\xi \,dx\\\\ &=2\pi \phi(-1) \end{align}$$

Hence, in distribution $e^{ix}=\delta(x+1)$.

Mark Viola
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It can be a bit confusing with the variables in the definition of the Fourier transform of a tempered distribution. We are used to having $x$ or $t$ as variable for the plain function, and $\xi$ or $\omega$ for the transformed function, but of course we can have any variable. Below I will use $y$ for the transformed function. Hopefully it helps you see what is going on.

Let $u$ be a tempered distribution. Its Fourier transform $\hat u$ is defined by $ \langle \hat{u}, \varphi \rangle = \langle u, \hat\varphi \rangle $ for all $\varphi\in\mathcal{S}.$

Abusing integral notation this means $$ \int \left( \int u(x) e^{-ixy} \, dx \right) \varphi(y) \, dy = \int u(x) \left( \int e^{-ixy} \varphi(y) \, dy \right) dx . $$

For $u(x) = e^{ix}$ this becomes $$ \langle \mathcal{F}\{e^{ix}\}, \varphi \rangle = \int \mathcal{F}\{e^{ix}\}(y) \, \varphi(y) \, dy = \int \left( \int e^{ix} e^{-ixy} \, dx \right) \varphi(y) \, dy \\ = \int e^{ix} \left( \int e^{-ixy} \varphi(y) \, dy \right) \, dx = \int e^{ix} \mathcal{F}\{\varphi\}(x) \, dx . $$

md2perpe
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