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For even weight $k$, complete the additive twist of a cusp-form L-function as $M(f,a/N,s)=(N/(2\pi ){)}^s\Gamma (s)L(f,a/N,s)$. Split its Mellin transform at $y=1/N$ and use rational-cusp inversion of a level-one cusp form in the lower integral. The resulting two integrals over $[1/N,\infty )$ converge locally uniformly for every complex $s$, giving an entire continuation and the displayed functional equation. The relation $ad\equiv 1(\mathrm{mod}N)$ identifies the opposite cusp.

The rational translate $g(\tau )=f(a/N+\tau )$ has Fourier coefficients $e^{2\pi ian/N}{c}_n$, with the same polynomial bound as $c_n$. Thus the preceding Mellin transform gives, initially for $\mathrm{Re}s$ large, the completion of the additive twist of a cusp-form L-functionWrite $\epsilon =i^k=(-1{)}^{k/2}$. Nonzero level-one forms have even weight, so ${\epsilon }^2=1$. Split the integral at $1/N$. In its lower part substitute $y=1/(N^{2}v)$, and use rational-cusp inversion of a level-one cusp form with $\tau =iv$:The bounds reverse from $(0,1/N)$ to $(\infty ,1/N)$, so

Both cusp values in this integral are $O(e^{-2\pi y})$ as $y\to \infty$. On any compact set of $s$-values, the remaining powers of $Ny$ are bounded by fixed positive powers of $y$; differentiation in $s$ merely introduces powers of $\log (Ny)$. The integral and its differentiated integrals therefore converge uniformly on compact sets. It defines an entire function of $s$, agreeing with the original completion in its half-plane of convergence, and hence supplies the required analytic continuation to $\mathbb{C}$.

For the opposite rational cusp, $a^{\prime }=-d$ has inverse $d^{\prime }=-a$ modulo $N$. Denote the two individual upper integrals by $I_a(s)$ and $I_{-d}(s)$. The representation givesReplacing $s$ by $k-s$ in the first equality and using ${\epsilon }^2=1$ proves the functional equation of an additive cusp-form twist:For $N=1$ this specializes to the entire completion and functional equation of the ordinary L-function of a cusp form.