= Solution
{c}
A <dominated differentiation under the integral sign> suffices. Suppose $\phi(\cdot,t)$ is measurable, $\phi(\cdot,t_0)$ is integrable at one parameter value, and for almost every $x$ the map $t\mapsto\phi(x,t)$ is continuously differentiable on an interval. If $|\partial_t\phi(x,t)|\leq H(x)$ throughout that interval for an integrable $H$, then the parameter integral is differentiable, and
$$
\frac d{dt}\int\phi(x,t)dx=\int\partial_t\phi(x,t)dx.
$$
One direct sufficient version assumes $\partial_t\phi(x,t)$ 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 $u\neq0$, the elementary bound $|e^{iux}-1|\leq|ux|$ gives
$$
\left|f(x)e^{itx}\frac{e^{iux}-1}{u}\right|
\leq|xf(x)|.
$$
The 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, proves
$$
\boxed{\widehat f'(t)=\int_{\mathbb R}ixf(x)e^{itx}dx.}
$$
This supplies the domination explicitly rather than merely differentiating a formal <Fourier transform>.
Back to article page
