Moving Differential Out of Integral of Fourier Transform Assume that a function $f$ is defined such that $xf\left(x\right)$ is integrable:
\begin{equation}
\int_{-\infty}^{\infty}\lvert xf\left(x\right)\rvert \mathrm{d}x < \infty.
\end{equation}
Then the function $xf\left(x\right)$ is of a Fourier transform, and so does $-2\pi i x f\left(x\right)$. The derivation is given by
\begin{equation}
\begin{aligned}
\mathcal{F}\left(-2\pi i x f\left(x\right)\right) &= \int_{-\infty}^{\infty}\left(-2\pi i x\right)e^{-2\pi i s x}f\left(x\right)\mathrm{d}x\\
&= \int_{-\infty}^{\infty}\left(\frac{\mathrm{d}}{\mathrm{d}s}e^{-2\pi i s x}\right)f\left(x\right)\mathrm{d}x\\
&\stackrel{1}{=}\frac{\mathrm{d}}{\mathrm{d}s}\int_{-\infty}^{\infty}e^{-2\pi i s x}f\left(x\right)\mathrm{d}x\\
&= \frac{\mathrm{d}}{\mathrm{d}s}\left(\mathcal{F}f\right)\left(s\right). 
\end{aligned}
\end{equation}
I am not sure why in step 1, the differential can be moved out. The reason given by the author is that $xf\left(x\right)$ is integrable. Can someone let me know more details about the argument? Also, it was mentioned that the derivative is continuous. Can someone also explain why the derivative is continuous?
 A: You need to know under what conditions
$$ \lim_{h\to 0} \int \frac{ f(x,s+h)-f(x,s)}h dx$$
exists. Suppose that $\partial_s f$ exists for a.e. $x$ and all $s$. Then Mean value theorem gives that there exists $t$ in between $s$ and $s+h$ such that
$$\frac{ f(x,s+h)-f(x,s)}h = \partial_s f(x,t(x,s))$$
So one way to proceed would be to ask for this derivative to satisfy a dominated convergence type estimate, uniform in the second parameter: that is, there exists $g\in L^1$ such that
$$|\partial_s f(x,s)|\le g(x)$$
for any $s$.
You should be able to check that the condition on $xf(x)$ allows you to apply this result
(In summary:  $f(x,s)\in L^1(dx)$ for all $s$, $\partial_s f(x,s)$ exists for a.e. $x$ and all $s$ with a dominating function lets you interchange $d/ds$ with $\int dx$.)

I did some more searching. Heres what I found:

*

*MSE posts; Difference of differentiation under integral sign between Lebesgue and Riemann  and  Lebesgue integral, differentiation under the integral sign . Also Differentiation under the integral sign using Fubini. for the Fubini argument mentioned by Saad.


*Planetmath page with a number of variants, including stronger ones that I did not previously know: it is allegedly enough for $f$ to be absolutely continuous in $s$ with an a.e. derivative in $L^1_{loc}(dxds)$. Unfortunately no proofs provided, except for Theorem 4 (distributional version) in the following PDF of the author "Steve Cheng" which I guess answers Weak derivative under the integral sign if someone wants to learn it and then write an answer :) (also posted on MO)


*The following paper discussing a Henstock-Kurzweil version of DUI which is apparently necessary and sufficient (at least for functions on the plane):
Talvila, Erik, Necessary and sufficient conditions for differentiating under the integral sign, Am. Math. Mon. 108, No. 6, 544-548 (2001). ZBL0990.26008.  (some cliffnotes/commentary)


*K. Conrad's examples (and non-examples)
I did search Royden and Axler but did not find this theorem. Wikipedia also has this theorem and proof but oddly, does not source it.
