 Solution

A dominated differentiation under the integral sign suffices. Suppose $\varphi (\cdot ,t)$ is measurable, $\varphi (\cdot ,t_0)$ is integrable at one parameter value, and for almost every $x$ the map $t\mapsto \varphi (x,t)$ is continuously differentiable on an interval. If $|{\partial }_t\varphi (x,t)|\le H(x)$ throughout that interval for an integrable $H$, then the parameter integral is differentiable, andOne direct sufficient version assumes ${\partial }_t\varphi (x,t)$ is continuous in $t$ almost everywhere and has that common domination; dominated convergence theorem then proves the assertion.

Here $|f(x)e^{itx}|=|f(x)|$, so the complex-valued function $h$ is integrable for every real $t$. For a real increment $u\ne 0$, the elementary bound $|e^{iux}-1|\le |ux|$ givesThe right side is integrable by hypothesis and the integrand converges pointwise to $ixf(x)e^{itx}$. dominated convergence theorem, applied to real and imaginary parts, provesThis supplies the domination explicitly rather than merely differentiating a formal Fourier transform.