= Solution
Put $g=f|_k[\alpha_N]$. Termwise <Mellin transform> in the initial half-plane gives
$$
I(f,s):=\int_0^\infty f(iy)y^{s-1}\,dy
=(2\pi)^{-s}\Gamma(s)L(f,s),
$$
so $\Lambda(f,s)=N^{s/2}I(f,s)$.
From the definition of the slash action,
$$
g(iy)=N^{-1}(iy)^{-k}f\left(\frac{i}{Ny}\right),
$$
or equivalently
$$
f\left(\frac{i}{Ny}\right)=Ni^ky^kg(iy).
$$
Substituting $y=1/(Nt)$ in the Mellin integral therefore gives
$$
I(f,s)=i^kN^{1-s}I(g,k-s).
$$
Multiplying by $N^{s/2}$ and recognizing the completed function on the right yields
$$
\boxed{\Lambda(f,s)=i^kN^{1-k/2}
\Lambda(f|_k[\alpha_N],k-s)}.
$$
At infinity, both $f(iy)$ and $g(iy)$ decay exponentially because they are <cusp forms>. The transformation just used converts the behavior of $f(iy)$ near zero into the exponential decay of $g(i/(Ny))$ at infinity. Consequently $I(f,s)$ converges absolutely for every $s\in\mathbb C$ after splitting the integral at one and applying that substitution to the part near zero. It is locally uniformly convergent in $s$, hence entire, and agrees with the original Dirichlet series in its initial domain. This proves the analytic continuation and the stated functional equation, as summarized by the <Mellin transform of a cusp-form L-function>.
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