Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 24 / 3

The analytic input for these Mellin transforms is elementary cusp decay. For a level-one weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(x+iy)|$ is invariant under the modular group. It is bounded on the standard fundamental region, because the cusp expansion decays exponentially, and hence bounded on all of $\mathbb{H}$. Fourier inversion givesTaking $y=1/n$ proves the Fourier coefficient bound for a cusp form $c_n=O(n^{k/2})$. In particular rational translates have polynomially bounded coefficients and exponential decay as $y\to \infty$, uniformly in their real parts. These facts justify the initial Mellin integrations and the later entire continuation.

Use $q=e^{2\pi i\tau }$, so $g(iy)={\sum }_{n\ge 1}{b}_n{e}^{-2\pi ny}$. Put $u=\mathrm{Re}s$. For $u>\max \{0,\sigma +1\}$,The Gamma integral and absolute convergence therefore justify exchanging summation and integration. For every positive integer $n$, substitution $t=2\pi ny$ givesSumming these identities proves the Mellin transform formula for the Dirichlet series:The power of the positive real variable $y$ uses the real logarithm; the initial half-plane stated above is a sufficient one even when $\sigma$ is negative.

Choose $b=(ad-1)/N\in \mathbb{Z}$, and take the matrixAt $w=\tau -d/N$ its denominator is $Nw+d=N\tau$, while its numerator is $aw+b=a\tau -1/N$. ConsequentlyApplying the weight-$k$ transformation law of the cusp form at $w\in \mathbb{H}$ proves the rational-cusp inversion of a level-one cusp form:

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.