# Show Smoothness by Morera

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.

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I tried to enlarge the fractional exponents (\large \frac ... ), but in integrals like this, one can always use $\exp{(\cdots)}$ = \exp{(...)} to help show where the action is taking place. –  amWhy Jan 14 '13 at 14:38
I believe you need to show the integral over an arbitrary smooth loop is zero, not just over any circle. Also, you might want to try switching the order of summation and integration. –  Potato Jan 14 '13 at 14:40
–  Willie Wong Jan 16 '13 at 12:42
I don't think that the hint "Morera" is very helpful. Just show that the series is uniformly convergent on any compact rectangle $[h, T]\times[-M,M]$, $\ h>0$. –  Christian Blatter Jan 16 '13 at 13:18
Thank you all for your responses. If we were able to show uniform convergence of the series on any compact rectangle $\Omega:=[h,T] x [-M,M], h>0,$ would the argument be, that since the series is smooth as a composition of smooth functions, that then f(t,x) is smooth on any $\Omega$ and we're therefore done? –  Damian Jan 16 '13 at 15:41