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

There is no singularity at the lower endpoint one. Put $t=\log (1+x)$, so $dx=e^{t}dt$. The nonnegative integral becomesFor $\alpha =1$, it converges as a reciprocal-square integral. For $\alpha >1$, exponential decay makes it converge. For $\alpha <1$, its integrand eventually grows without bound, so its integral diverges. ThusThis is the logarithmic borderline in a power-tail integrability test.

The function is measurable and $0\le {\cos }^2(1/x)\le 1$. Since $(0,1)$ has finite Lebesgue measure, comparison gives ${\int }_0^{1}g\le 1$. Therefore $g$ is integrable on $(0,1)$, despite oscillating infinitely often near zero.

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