Conditions for the differentiation of Fourier Transform of a function.

Question

Suppose $\varphi(t)$ is smooth on $\mathbb{R}$ and $f(x)=\displaystyle \int_{-\infty}^{\infty} \varphi(t) e^{-2\pi ixt}dt$ is the Fourier Transform of $\varphi(t)$. Then Inverse Fourier Transform is given by: \begin{equation} \varphi(t)=\displaystyle \int_{-\infty}^{\infty} f(x) e^{-2\pi ixt}dx \end{equation} My question is what should be the conditions on $f(x)$ so that I can differentiate $\varphi(t)$? Can I differentiate to get: \begin{equation*} \varphi'(t)=(-2\pi i)\displaystyle \int_{-\infty}^{\infty} xf(x) e^{-2\pi ixt}dx \end{equation*}

Answer 1

This is a fundamental theorem in the Lebesgue Theory of Integration :

Let ${\displaystyle X}$ be an open subset of ${\displaystyle \mathbf {R} }$, and ${\displaystyle \Omega }$ be a measure space. Suppose ${\displaystyle f\colon X\times \Omega \rightarrow \mathbf {R} }$ satisfies the following conditions:

${\displaystyle f(x,\omega )}$ is a Lebesgue-integrable function of ${\displaystyle \omega }$ for each ${\displaystyle x\in X}$. For almost all ${\displaystyle \omega \in \Omega }$, the derivative ${\displaystyle f_{x}}$ exists for all ${\displaystyle x\in X}$. There is an integrable function ${\displaystyle \theta \colon \Omega \rightarrow \mathbf {R} }$ such that ${\displaystyle |f_{x}(x,\omega )|\leq \theta (\omega )}$ for all ${\displaystyle x\in X}$ and almost every ${\displaystyle \omega \in \Omega }$. Then by the Dominated convergence theorem for all ${\displaystyle x\in X}$, ${\displaystyle {\frac {d}{dx}}\int _{\Omega }f(x,\omega )\,d\omega =\int _{\Omega }f_{x}(x,\omega )\,d\omega .}$