I am reading Iwaniec's Topics in Classical Automorphic Forms (this is coming from section 5.5). I am curious about the following statement. He states the following, I will note beforehand that $J_\ell(x)$ represents a Bessel function and $g\in C_0^\infty(0,\infty)$. Now he says the following:
We begin by applying the integral representation $$J_\ell(x)=\int_{-1/2}^{1/2}e(\ell t)e^{-ix\sin 2\pi t}dt$$. This can be regarded as the Fourier transform of the periodic function $\exp(-ix\sin 2\pi t)$; therefore, by Poisson's formula, $$ \sum_{\ell}g(\ell)J_\ell(x)=\int_{-\infty}^\infty\hat{g}(t)e^{-x\sin 2\pi t}dt $$ where $\hat{g}(t)$ is the Fourier transform $$ \hat{g}(t) =\int_{-\infty}^{\infty}g(y)e(ty)dy. $$
I will note that $e(x):=e^{2\pi i x}$. Now I am confused on the formula where he applies Poisson's formula. Namely, I am confused on where the integral comes in, as the version of Poisson summation I am used to is $\sum_n f(n)=\sum_n\hat{f}(n)$. Thus, I don't see how he gets an integral out of things. If there is something I'm missing or a reference that would be appreciated.
A friend helped me solve this, so I will type up the solution in case someone else comes upon this in the future.
The key is to use notice that $g$ is not periodic, so we shall define the following auxiliary function: $$ G(t)=\sum_{m}g(m)e(-tm). $$ This is a $1$-periodic function, and by the Poisson summation formula, we have that $$ G(t)=\sum_{m}\hat{g}(m+t). $$
Now for notation sake, we shall define $\phi_x(t)=\exp(-ix\sin 2\pi t)$, so that we have that $J_\ell(x)=\hat{\phi}_x(\ell)$ by the integral representation formula. Similarly, we shall see that $$ \hat{G}(\ell)=\int_{-1/2}^{1/2}G(t)e(\ell t)dt=\int_{-1/2}^{1/2}\sum_{m}g(m)e(-tm)e(\ell t)dt=g(\ell) $$
Now we will have that $$ \sum_{\ell}g(\ell)J_\ell(x)=\sum_{\ell}\hat{G}(\ell)\hat{\phi}_x(\ell) $$ This can be thought of as the inner-product of $\hat{G}(\ell)$ and $\hat{\phi}_x(\ell)$ in the space $\ell^2$. Now we note that if we have $f,g\in L^2(S^1)$, then we will have that $$ \int_{-1/2}^{1/2}f(t)g(t)dt=\sum_{\ell}\hat{f}(\ell)\hat{g}(\ell). $$ Applying this formula we have that $$ \sum_{\ell}\hat{G}(\ell)\hat{\phi}_x(\ell)=\int_{1/2}^{1/2}G(t)\phi_x(t)dt=\int_{-1/2}^{1/2}\sum_m\hat{g}(m+t)\phi_x(t)dt=\int_{-\infty}^\infty \hat{g}(t)\phi_x(t)dt $$ where the last equality comes from swapping the sum and the integral and realizing that we are now integrating over the entire real line. This is precisely what we set off to show.