Integrability of the Hilbert transform of a Schwartz function
$Hf$ does not have to decay faster than any polynomial; as @Andrew suggested, $1/x^2$ is the best we can expect.
The assumption $\int_{\mathbb R}f(t)\,dt=0$ implies that the antiderivative $F(t):=\int_{-\infty}^t f(s)\,ds$ is also integrable (in fact, it's in the Schwartz class too). This should allow you to integrate the second integral by parts (and the first one can be estimated with the MVT.)