Identical Summation and Integration of specific functions

A strange coincidence which I discovered recently, is that $$\int_{-\infty}^{\infty}{\tan^{-1}{\frac{1}{(x-\alpha)^2+\frac{3}{4}}}}dx = \sum_{x=-\infty}^{\infty}{\tan^{-1}{\frac{1}{(x-\alpha)^2+\frac{3}{4}}}}$$ Are there any other functions for which this characteristic holds true, and if so, what do they all have in common, and how can we determine them (if possible)?

• There exists $c>0$ with the following property: If $f$ is integrable on the line and the Fourier transform $\hat f$ is supported in $[-c,c]$ then $\int f=\sum f(n)$. This follows from the Poisson summation formula - I don't have time to get the $\pi$'s straight, maybe someone else will. – David C. Ullrich Aug 29 '15 at 3:29
• Where is $k$ in the rhs ? – Claude Leibovici Aug 29 '15 at 6:09
• Claude Leibovici, fixed the error, thanks for pointing it out. – Jack Lam Aug 29 '15 at 7:27

Say $\phi\in C^2_c(\Bbb R)$ (this is stronger than necessary) and the support of $\phi$ is contained in $[-\pi,\pi]$. Let $$f(x)=\frac1{\sqrt{2\pi}}\int e^{-itx}\phi(t)\,dt\quad(*).$$Then $$\int f(x)\,dx=\sum_{n\in\Bbb Z}f(n).$$First the inversion formula for the Fourier transform shows that $$\frac1{\sqrt{2\pi}}\int f(x)\,dx=\phi(0).$$But since $\phi$ is supported in the interval $[-\pi,\pi]$ it has a Fourier series; the $n$-th Fourier coefficient of $\phi$ is $\frac1{\sqrt{2\pi}}f(n)$, so we also have $$\frac1{\sqrt{2\pi}}\sum_{n\in\Bbb Z}f(n)=\phi(0).$$
Given $f$, how can you tell whether it arises from ($*$)? Arguing as above and adding a suitable version of the Paley-Wiener theorem shows this:
Theorem Suppose $f$ is an entire function in the complex plane, $|f(z)|\le ce^{\pi|z|}$, and $\int_{-\infty}^\infty|f(x)|\,dx<\infty$. Then $\int_{-\infty}^\infty f(x)\,dx=\sum_{n\in\Bbb Z}f(n)$.