As is well known the integral,
$$\frac{1}{2\pi} \int_{-T}^{T} e^{i\omega t} \, {\rm d}t = \frac{\sin(\omega T)}{\pi \omega}$$
converges to $\delta(\omega)$ as $T\rightarrow \infty$ in a distributional sense. As I think $\omega$ has to be real, how far is this representation valid in the sense of an "analytic continuation"?
Having a function $f(x)$ defined on the real axis which has a Taylor expansion, it is natural to get the analytic continutation by $x \rightarrow z=x+iy$. However in this case the integral is not convergent for complex $\omega=|\omega| \, e^{i\epsilon}$ with fixed $\epsilon$ as the RHS is badly divergent.
Still I sometimes see people in physics arguing by Wick rotation that the line $[-T,T]$ in,
$$\frac{1}{2\pi} \int_{-T}^{T} e^{i\omega t} \, {\rm d}t$$
can be rotated to complex $T$, e.g. the imaginary axis. But for such a rotation, the integral over the arc $[-T,-iT]$ would have to vanish, which does not happen.
Therefore in my opinion this is not possible, meaning that a distribution like that can not be analytically continued. Or does anyone have an idea what sense to make of $\delta(i\omega)$ for real $\omega$?