= Solution
With the <determinant-normalized slash operator>, let $W_N=\begin{pmatrix}0&-1\\N&0\end{pmatrix}$ and $Jf=i^k(f|_kW_N)=g$. 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 $J^2f=f$; 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 relations
$$
\boxed{G(y)=y^{-k}F(1/y),\qquad F(y)=y^{-k}G(1/y).}
$$
In 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-function>
$$
\Lambda(f,s)=\int_0^\infty F(y)y^{s-1}\,dy.
$$
The 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 gives
$$
\boxed{\Lambda(f,s)=\int_1^\infty\bigl(F(y)y^{s-1}+G(y)y^{k-s-1}\bigr)\,dy.}
$$
Applying the same formula to $g$ interchanges $F,G$, because $J^2=1$. It proves the <phase-normalized Fricke functional equation>
$$
\boxed{\Lambda(f,s)=\Lambda(g,k-s)\quad\text{for every }s\in\mathbb C.}
$$
The entire function here is the completion, despite the apparent <poles> of the gamma factor in its initial product formula.
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