TITLE: Solve $f(x)=\int_{x-1}^{x+1} f(t) \text{d}t$
QUESTION [11 upvotes]: Solve $f(x)=\int_{x-1}^{x+1} f(t) \text{d}t$. We know that $f(x)=0$ is a solution, are there any other solutions?
I suppose I can start with $f(x)=\frac{dg(x)}{dx}$, and then:
$$\begin{align*}
\frac{dg(x)}{dx}&=\int_{x-1}^{x+1} \frac{dg(t)}{dt} dt \\
\frac{dg(x)}{dx}&=g(x+1)-g(x-1) \\
\end{align*}$$
But then I'm stuck.

REPLY [8 votes]: A set of solutions is $$f(x) = a\sin(\omega x) + b\cos(\omega x)$$ for any $a,b\in\mathbb{R}$.  Note that $$\int_{x-1}^{x+1}f(t)\,dt = \frac{2\sin\omega}{\omega}\left(a\sin(\omega x)+b\cos(\omega x)\right)= \frac{2\sin\omega}{\omega}f(x)$$
so $\omega$ must be chosen to satisfy $$\frac{2\sin\omega}{\omega} = 1.$$  Numerically, $\omega \approx \pm 1.89549$.