Integral of $te^{-i\omega t}$ and Fubini's Theorem

Question

I have a question about the value of the integral $\int_0^\infty t e^{-i \omega t} dt$. I stumbled upon it while trying to compute the Fourier transform of $|t|$.

On one hand, we have that $|t e^{-i \omega t}|$ doesn't go to 0 as $t \to \infty$. This implies that the integral diverges.

On the other hand, using the trick of $\int_0^\infty e^{-xt} dx = 1$ for all $t$, we can write \begin{eqnarray*} \int_0^\infty t e^{-i \omega t} \times 1 dt & = & \int_0^\infty t e^{-i \omega t} \times \int_0^\infty e^{-xt} dx dt \\ & \substack{= \\ (1)}& \int_0^\infty \int_0^\infty t e^{-i \omega t} e^{-xt} dt dx \\ & = & \int_0^\infty \int_0^\infty \dfrac{e^{-t(i \omega +x)}}{i\omega + x} dt dx \\ & = & \int_0^\infty \dfrac{1}{(i\omega + x)^2} dx = \dfrac{1}{i \omega} \end{eqnarray*} where in $(1)$, we use Fubini's Theorem.

One of these two statements must be wrong, but which one ?

Thanks for your help.

Best, Olivier

Answer 1

Here is a graph for $$ {\rm Re} \left( \int _{0}^{S}\!t{{\rm e}^{it}}{dt} \right) $$ for $S$ from $0$ to $50$. (So take $\omega=-1$.) Does it look like it converges as $S \to \infty$?

Of course Fubini would only be applicable for a function with $\int_0^\infty \int_0^\infty |f(t,x)|dtdx < \infty$.