 Solution

With the determinant-normalized slash operator, let $W_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$ and $Jf=i^k(f{|}_k{W}_N)=g$. Conjugation by $W_N$ preserves ${\Gamma }_1(N)$, and rational slash operators preserve modular cusp holomorphy and vanishing. Hence $g$ is also a cusp form at that level. Direct substitution gives $J^{2}f=f$; the factor $i^k$ cancels the central weight sign.

Put $F(y)=f(iy/\sqrt{N})$ and $G(y)=g(iy/\sqrt{N})$. The defining formula gives the exact relationsIn the initial half-plane of absolute convergence, termwise integration of the Fourier series and the gamma function yield the Mellin transform of a cusp-form L-functionThe scaling of $y$ accounts for $N^{s/2}$ in the completion. At infinity $F$ and $G$ decay exponentially. At zero the boxed relation expresses $F$ as a power times an exponentially decaying function of $1/y$. Thus this integral converges locally uniformly for every complex $s$, including after differentiation in $s$, and defines an entire function.

Splitting at one and changing $y$ to $1/y$ in the lower integral givesApplying the same formula to $g$ interchanges $F,G$, because $J^2=1$. It proves the phase-normalized Fricke functional equationThe entire function here is the completion, despite the apparent poles of the gamma factor in its initial product formula.