Show this a tempered distribution I want to show that $f(x) : = e^{ie^x}$ is a tempered distribution. Therefore I need to show that for all $\varphi \in \mathscr{S}(\mathbb{R})$, that $$\int_{\mathbb{R}} \varphi(x) e^{ie^x}dx < \infty.$$ Equivalently, I could show that $$\int_{\mathbb{R}} \varphi(x) \mathcal{F}(e^{ie^x})dx < \infty,$$ where $\mathcal{F}$ denotes the Fourier transform. I have had no luck with this as of yet. 
 A: As Guy Fsone pointed out, it es enough to know that Schwartz-functions are integrable to prove that the integral against $e^{ie^x}$ is finite. The simple reason for that is just that $\vert e^{ie^x} \vert =1$ for all $x$.
On the other hand you need a little more. By showing that the integral is finite, you prove that integration against $e^{ie^x}$ is a well defined linear map $\mathcal{S}(\mathbb{R}) \rightarrow \mathbb{R}$. To prove that it is a tempered distribution it remains to show the continuity of that map. 
Equivalently you can prove a bound of the form
$$\left\vert \int \varphi(x) e^{ie^x} dx \right\vert \le C p_N(\varphi)$$
where $p_N(\phi) = \max_{k,l\le N} \vert\vert x^k \phi^{(l)} \vert\vert_{L^\infty}$ is one of the seminorms of $\mathcal{S}(\mathbb{R})$.
In order to get this bound observe that
$$ \left\vert \int \varphi(x) e^{ie^x} dx \right\vert \le \int \vert(1+x^2)\varphi(x)\vert \cdot (1+x^2)^{-1} dx \le \int (1+x^2)^{-1} dx  \cdot \sup_{x\in\mathbb{R}}\vert(1+x^2)\varphi(x)\vert $$
Now it is not to hard to see that the first factor is some finite number $C/2$ and the second factor is $\le 2 p_2(\varphi)$. This proves the statement. (The argument of course also works if you replace $f$ by any other bounded function.)
