Harmonic oscillator propagator in Euclidean time I'm following Nastase's book on Quantum Field Theory but this question is just about quantum mechanics in the path integral formalism. In chapter 8 he considers the propagator equation for a harmonic oscillator
$$\left(\frac{d^2}{dt^2}+\omega^2\right)\Delta(t-t')=\delta(t-t'),$$
which under Wick rotation $t\to -i\tau$ turns into
$$\left(-\frac{d^2}{d\tau^2}+\omega^2\right)\Delta_{E}(\tau-\tau')=\delta(\tau-\tau'),$$
where the subscript $E$ stands for Euclidean.
Now, I'm having troubles with checking that the unique periodic solution $$\Delta_{E}(\tau-\tau'\pm\beta)=\Delta_{E}(\tau-\tau')$$ for the propagator equation in Euclidean time turns to be
$$\Delta_E(\tau-\tau')=\frac{1}{2\omega}\left[\left(1+\frac{1}{e^{\beta|\omega|}-1}\right)e^{-\omega(\tau-\tau')}+\frac{1}{e^{\beta|\omega|}-1}e^{\omega(\tau-\tau')}\right].$$
When I treat this as an ansatz and plug it back in the propagator equation I don't get the $\delta(\tau-\tau')$ so I would appreciate any insight on this computation.
 A: This is the way I'd  do the Bosonic problem for the case $\beta=2\pi$, $\omega=M$. (I am cut and pasting  from old notes which is why I have chosen these parameters)
Star with
$$
\Delta(\tau-\tau')= \frac 1 {2\pi} \sum_{n=-\infty}^\infty \frac{e^{in(\tau-\tau')}}{n^2+M^2}
$$
which gives the delta via
$$\sum_{n=-\infty}^\infty e^{in\tau}= \sum_{m=-\infty}^\infty 2\pi \delta(\tau+2\pi m).$$
Now evaluate the sum as follows:
$$
 \frac 1{2\pi} \sum_{n=-\infty}^\infty \frac{e^{in\tau}}{n^2+M^2}=
   \sum_{n=-\infty}^\infty   \frac 1{2|M |} e^{-|M||\tau+2\pi n|}, \quad \hbox{(Poisson Summation)}\nonumber\\
   =  \frac 1 {2M} \frac{\cosh(\pi -\tau)M}{\sinh \pi M}, \quad0 <\tau<2\pi,\nonumber\\
   = \frac 1{2M} e^{-M\tau} +\frac 1 M\frac{ \cosh M\tau}{(e^{2\pi M}-1)}\quad0 <\tau<2\pi.\nonumber
  % = \frac 1{2M} (\coth \pi M\cosh M \tau- \sinh M\tau) \nonumber
 $$
The first line come from applying Poisson summation to the zero temperature expression
$$
\int_{-\infty}^{\infty} \frac{dk}{2\pi}\frac{e^{ik\tau}} {k^2+M^2}=\frac 1 {2|M|}e^{-|\tau||M|} 
$$
and has the  physical interpretation as  the method-of-images sum over the  $n$-fold winding of the particle trajectory around the periodic imaginary time direction.
The passage from the first  to second  lines is just summing  the two  geometric series from  $n=0$ to $ \infty$ and $n=-\infty$ to $-1$.
In this "$\cosh(\pi -\tau)$" version  the delta function comes from the restriction on $\tau$ which leads to discontinuity in the slope of $\Delta(\tau-\tau')$  at $\tau-\tau'=2\pi m$.
