Past exam of the options I course of the University of Oxford / 2010 / D / D2 / 2 / Solution

For $x\ge 1$, $|\sin (1/\sqrt{x})|\le 1/\sqrt{x}$, soThus $g$ is integrable on this interval. But for $x>1$, $\sqrt{x^2-1}<x$, hence $h(x)>1/x$. The integral of $1/x$ diverges at infinity, so $h$ is not integrable. Defining $h(1)=0$ changes only a null set and has no effect on this conclusion.