Taking the limit in Poisson summation formula as the step size tends to zero I'm working on the latter half of Exercise 9.11 from Rudin's Real and Complex Analysis about the Poisson summation formula. Let $\alpha$ and $\beta$ be positive numbers such that $\alpha\beta = 2\pi$. For a sufficiently "nice" function $f$ on the real line,
$$ \sum_{k=-\infty}^{\infty} f(k\beta) = \frac{\alpha}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty} \hat{f}(n\alpha), \tag{1} $$
where $\hat{f}$ is the (scaled) Fourier transform of $f$:
$$ \hat{f}(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(t)e^{-ixt} \:dt. $$
The problem is to describe what the Poisson summation formula entails by allowing $\alpha$ to approach $0$. I have a vague sense that the left side of (1) will tend to $f(0)$ and the right will tend to the scaled integral
$$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat{f}(x) \:dx, $$
provided $f$ or $\hat{f}$ are suitably nice, but I'm having a difficult time coming up with a rigorous description.
For the right side of (1), my guess comes from pulling the limit through the integral in the end:
$$ \lim_{\alpha\to 0} \frac{\alpha}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty} \hat{f}(n\alpha) = \lim_{\alpha\to 0} \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty} \int_{-\infty}^{\infty} \hat{f}(n\alpha)\chi_{[n\alpha,(n+1)\alpha)}(x) \:dx, $$
but I'm not sure what sensible conditions to put on $f$ to do so. Saying the family of integrands is dominated seems too artificial.
 A: Your idea is correct. The expression
$$\alpha \sum_{n=-\infty}^{\infty} \hat{f}(n\alpha)$$
is like Riemann sum for $\int_{-\infty}^\infty \hat f(\xi)\,d\xi$, except, of course, Riemann sums are normally considered on a finite interval. Dealing within this kind of sum is a bit annoying, so let's focus on the left hand side of $(1)$ instead. Suppose there are constants $C$ and $p>1$ such that
$$
|f(x)|\le C|x|^{-p}
$$
for large $x$. This isn't a super strong assumption; reasonable integrable functions tend to decay like that. 
As $\beta\to\infty$, we get
$$
\left| \sum_{k\ne 0} f(k\beta) \right| \le \beta^{-p}\sum_{k\ne 0}|k|^{-p} \to 0
$$
so the left hand side of $(1)$ indeed converges to $f(0)$. 
Of course, so does the right hand side since they are equal. At this point I'm inclined to cop out by saying: suppose also that $\hat f$ is integrable; then the Fourier inversion formula holds.
$$
f(0) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty\hat f(\xi)\,d\xi
$$ 
So that's the proof that 
$$\alpha \sum_{n=-\infty}^{\infty} \hat{f}(n\alpha)\to \int_{-\infty}^\infty \hat f(\xi)\,d\xi\tag2$$
as $\alpha\to 0 $.  

Proving $(2)$ directly seems awkward. 
If $f$ is integrable, then $\hat f$ is uniformly continuous on $\mathbb{R}$. 
which tells us that $\alpha \hat f(n\alpha)$ is uniformly close to $\int_{n\alpha}^{(n+1)\alpha}\hat f(\xi)\,d\xi$. Unfortunately, a uniform bound isn't much good with infinitely many terms. We need something like 
$$\left|\alpha \hat f(n\alpha) - \int_{n\alpha}^{(n+1)\alpha}\hat f(\xi)\,d\xi\right| < \frac{C(\alpha)}{n^p} \tag3$$
with $p>1$, and $C(\alpha)\to 0$ as $\alpha\to 0$; this way the errors add up to a small quantity. By the mean value theorem the left hand side of $(3)$ is bounded by $\alpha^2 \sup_{[n\alpha, (n+1)\alpha]}|\hat f'|$ so if we have $|\hat f'|\le C/|x|^p$ for some $1<p<2$, then $(2)$ follows. This is a relatively strong assumption compared to the first part of this post.
A: I think I have an extension to Michelle's answer above that proves (2) under different restrictions on $\hat{f}$. Suppose the principal value integral of $\hat{f}$ exists, that is, suppose
$$
\lim_{A\to\infty} \int_{-A}^{A} \hat{f}(x) \:dx < \infty,
$$
call this limit $I$, and suppose there exist constants $B$ and $p$, where $B<\infty$ and $p>1$, such that
$$
\lvert x^{p}\hat{f}(x)\rvert \leq B
$$
for every (sufficiently large) real number $x$. Let $\epsilon$ be a positive number, and let $A$ be a positive number such that
$$
\Bigl\lvert I - \int_{-A}^{A} \hat{f}(x) \:dx\Bigr\rvert < \epsilon, \quad
\int_{A}^{\infty} Bx^{-p} \:dx < \epsilon.
$$
Let $\alpha$ be a positive number such that $A/\alpha$ is an integer $K$, but small enough so that $\alpha\hat{f}(0)<\epsilon$ and
$$
\Bigl\lvert \int_{-A}^{A} \hat{f}(x) \:dx + \alpha\hat{f}(0) - \sum_{-K}^{K} \alpha \hat{f}(n\alpha)\Bigr\rvert < \epsilon.
$$
The possibility of this inequality follows from the assumption that $\hat{f}$ is Riemann integrable on $[-A,A]$. The Riemann sum takes the right endpoints as approximations on $[0,A]$ and left endpoints as approximations on $[-A,0]$.
The decay condition entails
$$
\Bigl\lvert \sum_{K+1}^{\infty} \alpha\hat{f}(n\alpha)\Bigr\rvert
 \leq \sum_{K+1}^{\infty} \alpha B(n\alpha)^{-p} \leq \int_{A}^{\infty} Bx^{-p} \:dx
 < \epsilon,
$$
since the middle expression is an infinite lower Riemann sum of the right integral using right endpoints, and an analogous statement holds for the lower tail of the series $\sum \alpha\hat{f}(n\alpha)$. Therefore,
$$
\Bigl\lvert I - \sum_{-\infty}^{\infty} \alpha\hat{f}(n\alpha)\Bigr\rvert \leq 5\epsilon.
$$
The crux of the argument is to use the decay to obtain a lower approximation for the tails.
