I'm trying to show smoothness on $(0,\infty)(\Re)$ of the following function:
$$ f(t,x)=\sum_{n=-\infty}^\infty e^{-\large \frac{(x-2\pi n)^2}{2t}}\frac{1}{\sqrt{2\pi t}} $$
The function is continous as composition of continous functions. The hint says, we should use Morera's Theorem and $$ \left\lvert e^{\large\frac{-(x+re^{i\theta}-2\pi n)^2}{2t}}\right\rvert = e^{\large\frac{r^2\sin(\theta)^2-(x+r\cos(\theta))^2}{2t}}e^{\large 2\pi n\frac{(x+r\cos(\theta))-\pi n}{t} }$$
That is why I want to show that $ \int_{\gamma} f(t,x) ds=0 $ for every closed curve in order to get smoothness through the Morera theorem.
I'd like to use that hint, that's why I chose an arbitrary circle around any x with radius r.
$$\int_{\theta=0}^{2\pi} f d\theta = 0 \Rightarrow \int_{\theta=0}^{2\pi} | f| d\theta = 0 \Rightarrow$$
$$ \frac{1}{\sqrt{2\pi t}}\int_{\theta=0}^{2\pi}\sum_{n=-\infty}^\infty \left\lvert e^{\large\frac{-(x+re^{\large i\theta}-2\pi n)^2}{2t}}\right\rvert d\theta = \frac{1}{\sqrt{2\pi t}}\int_{\theta=0}^{2\pi}\sum_{n=-\infty}^\infty e^{\large \frac{r^2\sin(\theta)^2-(x+r\cos(\theta))^2}{2t}}e^{2\pi n\large\frac{(x+r\cos(\theta))-\pi n}{t} } d\theta $$
Which is where I am stuck and don't know how to show that this Integral is zero. Any help is highly appreciated.