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
On
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 $$ \langle \mathcal{F}\{e^{ix}\}, \varphi \rangle = \int \mathcal{F}\{e^{ix}\}(y) \, \varphi(y) \, dy = \int \left( \int e^{ix} e^{-ixy} \, dx \right) \varphi(y) \, dy \\ = \int e^{ix} \left( \int e^{-ixy} \varphi(y) \, dy \right) \, dx = \int e^{ix} \mathcal{F}\{\varphi\}(x) \, dx . $$
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 $e^{ix}=\delta(x+1)$.