Mellin transform of a theta series and the zeta function

Example

For $x > 0$ let $\psi(x) = \sum_{n=1}^{\infty} e^{-\pi n^2 x}$. With the theta function $\theta(x) = \sum_{n \in \Z} e^{-\pi n^2 x}$, one has $\psi(x) = (\theta(x) - 1)/2$. For $\operatorname{Re} s > 1$, the Mellin transform of $\psi$ at $s/2$ is $$\int_0^\infty x^{s/2 - 1} \psi(x) \, dx = \pi^{-s/2} \, \Gamma\left(\frac{s}{2}\right) \zeta(s),$$ where $\Gamma$ is the gamma function and $\zeta$ is the Riemann zeta function.

Check: for each $n \ge 1$ the dilation rule gives $$\int_0^\infty x^{s/2 - 1} e^{-\pi n^2 x} \, dx = (\pi n^2)^{-s/2} \, \Gamma\left(\frac{s}{2}\right) = \pi^{-s/2} \, \Gamma\left(\frac{s}{2}\right) n^{-s}.$$ Let $\sigma = \operatorname{Re} s > 1$. The sum over $n$ of the integrals of the absolute values is $\pi^{-\sigma/2} \Gamma(\sigma/2) \sum_n n^{-\sigma} < \infty$. So, by Fubini's theorem, the sum and the integral may be exchanged, and the sum of the right sides is $\pi^{-s/2} \Gamma(s/2) \zeta(s)$.

The series $\psi$ is Mellin convergent at $s/2$ exactly for $\operatorname{Re} s > 1$. Near $\infty$, $\psi(x) \le C e^{-\pi x}$; near $0$, $\psi(x) \sim \frac{1}{2} x^{-1/2}$, which follows from the identity $\theta(1/x) = \sqrt{x}\, \theta(x)$.