Conditions for the differentiation of Fourier Transform of a function. 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*}
 A: 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 ${\displaystyle x\in X}$,
${\displaystyle {\frac {d}{dx}}\int _{\Omega }f(x,\omega )\,d\omega =\int _{\Omega }f_{x}(x,\omega )\,d\omega .}$ 
