 Solution

A dominated differentiation theorem is the following. Suppose $\varphi (x,t)$ is measurable in $x$ and differentiable in $t$ on an open interval, is integrable for at least one parameter value, and its parameter derivative is dominated in absolute value by an integrable function independent of $t$. Then its integral is differentiable andLocal versions require the domination only on a neighbourhood of each parameter. The proof bounds the difference quotients by the dominating function, using the fundamental theorem of calculus or the mean-value theorem, then applies dominated convergence. Apply the argument to real and imaginary parts for a complex integrand.

For fixed $t$, $f(x)e^{itx}$ is measurable and has absolute value $|f(x)|$, so is integrable. Its derivative with respect to $t$ is $ixf(x)e^{itx}$, whose absolute value is the assumed integrable function $|xf(x)|$. HenceDominated convergence also shows this derivative is continuous, though only differentiability is required.