Edit: I have found a reliable source which confirms my thoughts (http://www.tcm.phy.cam.ac.uk/~bds10/tp3/ans3.ps), however, no explicit calculation is given. Any help is appreciated.
This question is motivated by a physical problem, and I personally have not studied any distribution theory, so if anything I write is not completely rigorous, forgive/correct me.
Consider \begin{equation} U(q_a,q_b;T)=\sqrt{\frac{m\omega}{2\pi i\sin(\omega T)}}\exp\left(\frac{im\omega}{2\sin(\omega T)}((q^2_a+q^2_b)\cos(\omega T)-2q_aq_b))\right) \end{equation} as a function of $T$, where $q_a,q_b\in\mathbb{R}$. Usually, $T\ge0$ holds.
(Physically, this is the amplitude for a particle in a harmonic oscillator potential to go from $q_a$ to $q_b$ in a time $T$).
Now using the identity (stolen from http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/09/) \begin{equation} \delta(x)=\lim_{\epsilon\downarrow0}\frac{1}{\sqrt{2\pi i\epsilon}}e^{\frac{ix^2}{2\epsilon}}, \end{equation} one can easily show that for any $n\in\mathbb{N_{\ge0}}$ it holds that \begin{equation} \lim_{\epsilon\downarrow0}U\left(q_a,q_b;\frac{2n\pi}{\omega}+\epsilon\right)=\delta(q_a-q_b), \end{equation}
(This makes sense physically since after $2n$ times the period the particle is back where it was, so to speak). Similarly, we should have something like \begin{equation} \lim_{T\to\frac{(2n+1)\pi}{\omega}}U(q_a,q_b;T)=\delta(q_a+q_b), \end{equation} (which physically corresponds to the fact that after n times half a period the particle is at $-q_a$, i.e. on the other side of the harmonic well). However, I cannot seem to derive this. Namely, letting $n\in\mathbb{N}_{\ge0}$ and $\epsilon>0$, we have \begin{equation} \lim_{\epsilon\downarrow0}U(q_a,q_b;\frac{(2n+1)\pi}{\omega}\pm\epsilon)=\lim_{\epsilon\downarrow0}\sqrt{\frac{m}{\mp2\pi i\epsilon}}\exp\left(\frac{im}{\pm\epsilon}(q_a+q_b)^2)\right), \end{equation} which I cannot relate to this limit of the Delta distribution. A similar problem arises when trying to calculate \begin{equation} \lim_{\epsilon\downarrow0}U(q_a,q_b;\frac{2n\pi}{\omega}-\epsilon) \end{equation} for any $n\in\mathbb{N}_{\ge1}$, which should also give $\delta(q_a-q_b)$, but so far in my calculations it does not. Any hints/thoughts or comments are welcome.