Is it possible to have $\int_0^\infty f(x)\,dx$ convergent and $|f|=1$ everywhere? I came across a problem that asked if it is posible for a function to be Riemann integrable function in $[0,+\infty)$ but also $|f(x)|\geq 1$ for all $x\geq 0$. 
At first I thought it was imposible, but I realized that only holds for continuous functions, because they would have to be either positive or negative, and then they would have to go to 0 at infinity. 
I have an idea of what the function would have to be like, with alternating signs, but whose integral converges, but I haven't been able to find any, so I'm starting to think it is imposible. 
I would like some help finding this function, or disproving it, as I don't know many tools for working with functions without a constant sign.

The Fresnel integral:
$$\int_0^\infty e^{ix^2}dx=\frac{(1+i)\sqrt{2\pi}}{4}$$
(The problem statement doesn't say $f$ has to be real!)
