Does the integral $\tilde{G}(\omega)=\frac{1}{\omega_0}\int\limits_{-\infty}^{\infty}\sin(\omega_0t)\Theta(t)e^{i\omega t}dt$ exist for real $\omega?$

Question

Consider the causal, retarded Green's function of the undamped but forced harmonic oscillator $$G(t)=\frac{1}{\omega_0}\sin(\omega_0t)\Theta(t)$$ where $\omega_0$ is a real constant and $t$ is a real variable. Now, consider the Fourier transform $$\tilde{G}(\omega)=\int\limits_{-\infty}^{\infty}G(t)e^{i\omega t}dt=\frac{1}{\omega_0}\int\limits_{-\infty}^{\infty}\sin(\omega_0t)\Theta(t)e^{i\omega t}dt.$$

The integral above exists for complex $\omega$ with ${\rm Im}(\omega)>0$. However, my question is, does the integral $\tilde{G}(\omega)$ also exist for real values of $\omega$? If so, what is the value of $\tilde{G}(\omega)$?

Answer 1

It depends on the function $\Theta$, the problem is that the $e^{i\omega t}$ does not decay with time if $\Im(\omega)>0$ is not true. This is why for the fourier of some function: $$F(\omega)=\mathscr{F}[f(t)]\{\omega\}=\int\limits_{-\infty}^\infty f(t)e^{i\omega t}dt$$ to exist the conditions: $$f(t),f'(t)\to 0\,\,\,\,\,|t|\to\infty$$ are often used or multiplied by a fixed length pulse or given a decaying part to the function.

For some that appear to diverge you can use complex analysis and find a "Principle value" so that is also an option