First, note that the notion of “no pole” needs to be made very specific: the $j$-invariant map has no pole on $\mathbb{H}$ and is automorphic, yet is not constant. Of course, it has a “pole at infinity”.
If we want to split hairs, we could then consider $\exp{j}$, which does not have a pole at infinity, rather an essential singularity.
We actually want not to consider any of these somewhat pathological situations. Here’s how the classical approach goes.
Because $f(z)=f(z+1)$, you can write $f(\tau)=g(e^{2i\pi \tau})$ for some holomorphic function on the punctured open unit disk $g$.
In particular, $g$ has a Laurent series $g(q)=\sum_{n \in \mathbb{Z}}{a_nq^n}$.
We assume now that $g$ is holomorphic at zero ($g(0)$ corresponds to $f(i\infty)$ – if such a thing could be given sense – so we usually say that $f$ is holomorphic at infinity or that $f$ has no pole at infinity. In other words, the “no pole” assumption has to include infinity).
So $g(q)=\sum_{n \geq 0}{a_nq^n}$ is holomorphic on the full open unit disk.
This implies in particular that $f(z)$ is bounded on the subspace $S$ defined by $Im(z) \geq 1/2$ (ie $|e^{2i\pi z}| \leq e^{-\pi}$).
It’s an elementary exercise that every point of $\mathbb{H}$ has a $SL_2(\mathbb{Z})$-translate in $S$: in other words, every value taken by $g$ is in $g(\overline{D}(0,e^{-\pi}))$.
So the local maximum of $g$ on that disk is a global maximum on the open disk $D(0,1)$. By the strong maximum principle, $g$ is constant, hence so is $f$.
Best Answer
A change of variable should work \begin{align} \int_0^\infty e^{-\pi tn^2}t^{(s+\varepsilon)/2-1} dt &=_{u=\pi n^2t} \int_0^\infty e^{-u}\left(\frac{u}{\pi n^2}\right)^{(s+\varepsilon)/2-1}\frac{du}{\pi n^2} \\ =& \pi^{-(s+\varepsilon)/2} \Gamma\left(\frac{s+\varepsilon}{2}\right)n^{-s-\varepsilon} \end{align}