$f(t) = \cos t^{-1} + \int_t^\infty \frac{1}{\tau^2 + f(\tau)^2} d\tau$ implies the integral is $O(\frac{1}{t})$ The following is a quote from "asymptotic methods in analysis" by de Bruijn (p. 136).

If we know that the real function $f(t)$ satisfies the relation
  $$f(t) = \cos t^{-1} + \int_t^\infty \frac{1}{\tau^2 + f(\tau)^2} d\tau \,\,\,\,\,\,\,\,\,(t > 1)$$
  then it is easily seen that the integral is $O(t^{-1})$.

The $O$ notation is for $t \to \infty$. If $|f(t)| < t$, I can prove the integral is $O(1)$ by factoring out $1/\tau^2$ from the integrand and expanding the remaining factor using the binomial theorem. However, the author does not specify any extra condition on $f(t)$ at all. How can I prove it generally without the condition that $|f(t)| < t$?

Since $f$ is real valued, we have, for $0 < t \leqslant \tau$,
$$0 < \tau^2 \leqslant \tau^2 + f(\tau)^2,$$
and therefore
$$0 < \int_t^\infty \frac{d\tau}{\tau^2 + f(\tau)^2} \leqslant \int_t^\infty \frac{d\tau}{\tau^2} = \frac{1}{t}.$$
