I'm trying to understand the proof of the functional equation for the L-series of primitive, even Dirichlet characters. For even, primitive characters we have $$\theta_\chi(x):=\sum_{n\in \mathbb{Z}} \chi(n)\exp\left(\frac{-\pi}{q}n^2x^2\right).$$
In my lecture notes it says
The theta function decays exponentially as $x \rightarrow \infty$ [$\theta_\chi(x)=O(e^{-\pi/qx^2})$]
which I assume means $\theta_\chi(x)=O\left(\exp\left(\frac{-\pi}{q}x^2\right)\right)$ since I want to deduce that the integral
$$\int_1^\infty \theta_\chi(x)x^s\frac{dx}{x} $$
converges.
I "feel" like this is vaguely because when $n$ gets big, the $\exp\left(\frac{-\pi}{q}n^2x^2\right)$ is so small that is contributes basically nothing to the sum. But I don't really think this is enough, since
$$\int_{-\infty}^\infty \exp\left(\frac{-\pi}{q}x^2\xi^2\right)d\xi \leq \sum_{n \in \mathbb{Z}}\exp\left(\frac{-\pi}{q}n^2x^2\right) \leq 2+ \int_{-\infty}^\infty \exp\left(\frac{-\pi}{q}x^2\xi^2\right)d\xi$$
so
$$\frac{\sqrt{q}}{|x|} \leq \sum_{n \in \mathbb{Z}}\exp\left(\frac{-\pi}{q}n^2x^2\right) \leq 2+\frac{\sqrt{q}}{|x|}$$
which is not strong enough. So how do I prove that
The theta function decays exponentially as $x \rightarrow \infty$ [$\theta_\chi(x)=O(e^{-\pi/qx^2})$]
So, using Sanchez's hint, I think I have
$$\begin{array}{rll}\theta_\chi(x) & \leq & 2\exp\left(\frac{-\pi}{q}x^2\right)+\sum_{|n|\geq 2}\exp\left( \frac{-\pi}{q}n^2x^2\right) \\ & \leq & 2\exp\left(\frac{-\pi}{q}x^2\right)+\sum_{|n|\geq 4}\exp\left( \frac{-\pi}{q}n x^2\right) \\ \text{(expanding the sum} \\ \text{ as a geometric series)} & \leq &2 \exp\left(\frac{-\pi}{q}x^2\right)+ 2\exp \left(\frac{-\pi}{q}4x^2\right)\left( 1-\exp\left(\frac{-\pi}{q} x^2\right)\right)^{-1}\end{array}$$
which does it.