Referencing Don Zagier's notes (Utrecht), and also paper for moderate growth version.
The Rankin-Selberg transform can be defined by $\mathcal{R}(f,s) := \int_0^\infty \tau_2^{s-2} a_0(\tau_2) d\tau_2 = \mathcal{M}(\tau_2^{-1} a_0,s)$ (where $\mathcal{M}$ is the Mellin transform). By the main theorem, this has a simple pole at $s=1$ with residue $\kappa = \frac{\pi}{3}\int_{\mathcal{F}}f d\mu$, and elsewhere poles at $\rho/2$ where the $\rho$s are non-trivial zeroes of the Riemann zeta function.
Suppose that the constant term $a_0(\tau_2) = C_0 \tau_2^{\lambda_0}+C_1 \tau_2^{\lambda_1}+\cdots$ as $\tau_2 \rightarrow 0$. Then the Rankin-Selberg transform gives poles at $s=1-\lambda_j$ with residue $C_j$. Zagier writes that this suggests but unfortunately does not seem to imply that we have $a_0(\tau_2) \sim \kappa + \sum_{\rho: \zeta(\rho) = 0}C_\rho \tau_2^{1-\rho/2}$ as $\tau_2 \rightarrow 0$.
He continues, "under some circumstances it may be possible to prove this" (paraphrasing). Does anyone know references, or a direct answer to what one needs to prove this?