Lemma: Closed weight and its boundary control

Set

and

Then

At a generic angle this extends continuously to $u=0$. For $\beta =0$,

The function ${\Phi }_{\beta }$ is jointly locally integrable in $(u,\vartheta )$ up to $u=0$, including image boundaries, and decays exponentially as $u\to \infty$.

Proof. Split the series into $n\ge 0$ and $n\le -1$, and write $\sin x=(e^{\mathrm{i}x}-e^{-\mathrm{i}x})/(2\mathrm{i})$. The resulting two geometric sums give (eq:Phi-closed); the $n=0,\beta =0$ contribution is zero. Combining conjugate geometric denominators gives (eq:Carslaw).

The possible boundary poles are precisely $\vartheta \pm \pi \in 2\pi \alpha \mathbb{Z}$. Near each such angle ${\vartheta }_j$, the closed form is bounded by a constant times

This is locally integrable in the two variables. Away from those angles, all denominators remain nonzero at $u=0$.

For $0<\beta <1$, exponential decay follows with rate at least $\min (\beta ,1-\beta )/\alpha$. For $\beta =0$, the zero mode does not contribute, and a rate $1/\alpha$ suffices.