First, let me mention that this post is based on a previous post: https://physics.stackexchange.com/questions/782017/normalization-of-the-harmonic-oscillator-propagator/782035?noredirect=1#comment1756546_782035
I would like to understand better what is the necessary condition to use the simplification $$(1)\quad f(x)\,\delta(x-x_{0})\,=\,f(x_{0})\,\delta(x-x_{0}).$$ In particular, what are the necessary condition on $\Delta t$ in order to be able to identify the integral $$I\,=\,\frac{m\omega}{2\pi\sin(\omega\Delta t)}\int_{-\infty}^{\infty}dx_{1}\exp\left(\frac{im\omega^{2}}{2\sin(\omega\Delta t)}\left[(x_{0}^{2}-x^{2})\cos(\omega\Delta t)+2x_{1}(x-x_{0})\right]\right)$$ with a value of $I=\delta(x-x_{0})$? The current approach use the simplification
$$\frac{m\omega}{2\pi\sin(\omega\Delta t)}\int_{-\infty}^{\infty}dx_{1}\exp\left(\frac{im\omega^{2}}{\sin(\omega\Delta t)}x_{1}(x-x_{0})\right)=\delta(x-x_{0})$$ And then $$I\,=\,\exp\left(\frac{im\omega^{2}}{2\sin(\omega\Delta t)}\left[(x_{0}^{2}-x^{2})\cos(\omega\Delta t)\right]\right)\delta(x-x_{0})=\delta(x-x_{0})$$ based on (1). However, I cannot see how this can be done for the singular points in which $\sin(\omega\Delta t)=0$. Would I be correct to say that at these point the above argument fails?