{c}

For $x>1$, $1/x\in(0,1)$ and $\cos^2(1/x)\geq\cos^2(1)>0$. Its integral over the infinite interval is bounded below by the integral of a positive constant. Therefore \b[$g$ is not integrable on $(1,\infty)$]. In fact it tends to one at infinity.


Solution

Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / D / 2 / b / ii / Solution

 Solution (source code)