 Solution

Put $g=f{|}_k[{\alpha }_N]$. Termwise Mellin transform in the initial half-plane givesso $\Lambda (f,s)=N^{s/2}I(f,s)$.

From the definition of the slash action,or equivalentlySubstituting $y=1/(Nt)$ in the Mellin integral therefore givesMultiplying by $N^{s/2}$ and recognizing the completed function on the right yields

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.