# Fourier transform of the identity function $f(x)=x$

Let's say you are given $\omega_f \in \mathcal{S}'(\mathbb R)$ with \begin{align*} f \colon \mathbb{R} &\to \mathbb{R}\\ x &\mapsto x, \end{align*} and the definition of Fourier transform

\begin{align*} \hat{\phi}(\omega)=\int_{-\infty}^{+\infty}e^{-i\omega x}f(x)dx, \, \omega \in \mathbb{R}. \end{align*} Obviously, that function has no Fourier transform, but its corresponding tempered distribution $\omega_f$ does. Using \begin{align*} \widehat{ (\partial_j \phi)} (\omega) =& \,\, i\omega_j\hat{\phi}(\omega)\,\,\,\,\,\,\,\,\,\, (\star) \\ \widehat{1} =& \,\, 2\pi \delta(\omega) \end{align*} we find $$\widehat{ (\partial_jf)}(\omega)= \widehat{1}\Rightarrow\hat{f}(\omega)=-2 \pi i \frac{\delta(\omega)}{\omega}.$$ On the other hand, I know that \begin{align*} 2\pi\delta(\omega)=\int_{-\infty}^{\infty}e^{-i\omega x}dx \Rightarrow 2\pi \delta'(\omega)=-i\underbrace{\int_{-\infty}^{\infty}xe^{-i\omega x}dx}_{=:\,\hat{f}(\omega)} \end{align*} which gives $$\hat{f}(\omega) = 2\pi i\delta'(\omega).$$

Why do this two results contradict eachother?

Is it because ($\star$) is not true for $\phi \in \mathcal{S}'(\mathbb R^n)$?

Writing $\frac{\delta(\omega)}{\omega}$ is not a good idea, because strictly speaking, it's not well defined in many cases. But we do have the relation $\omega\delta'(\omega)=-\delta(\omega)$, so in this sense, your two answers are the same, they are both correct.