Context is I'm trying to learn differentiating under the integral sign and characteristic functions of random variables with little knowledge of complex analysis (long story).
Suppose $X$ is a random variable s.t. $E[|X|] < \infty$ and that $h(t,x)=ixe^{itx}$. Let $t \in [a,b] \subseteq \mathbb R$. Prove that
$$|iXe^{itX}|\le |X|$$
Here is what I tried.
$$|iXe^{itX}| = |i| |X| |e^{itX}| = |X| |e^{itX}| = |X| |e^{itX}| = |X| |\cos(tX)+i\sin(tX)| = |X|\sqrt{\cos^2(tX)+\sin^2(tX)} = |X|$$
Is there any $=$ that is wrong and actually supposed to be $\le$?
It is exactly as you say, if $t$ and $X$ are real-valued. But if $X$ isn't, the inequality doesn't hold either: if $X=Y+iZ$, with $Y,Z$ real, then $$ |e^{itX}|=|e^{itY}\,e^{-tZ}|=e^{-tZ}. $$