Past exam of the integration course of the University of Oxford / 2017 / 2 / a

Let $\varphi :I\to J$ be a $C^1$ diffeomorphism between intervals, which may be unbounded. For a nonnegative Lebesgue measurable function $f$ on $J$, Lebesgue substitution on an interval givesallowing the value infinity. The same identity applied to $|f|$ shows that $f$ is Lebesgue integrable exactly when the transformed integrand is integrable; in that case the displayed formula holds for signed or complex $f$ as well. The absolute value on the derivative makes the formula valid for both increasing and decreasing substitutions. Finite endpoints can be omitted without changing a Lebesgue integral, because individual points are null sets.

Both functions are continuous on the positive domains in question, so they are measurable. For the first function, the decreasing substitution $t=1/x$ and $|dx/dt|=t^{-2}$ giveFor every integer $k\ge 1$,Summing this lower bound gives a divergent harmonic series. Thus $f$ is not Lebesgue integrable on $(0,1)$.

For the second function, $|\sin x|\le x$ on $(0,1)$ and $|\sin x|\le 1$ everywhere giveTherefore $g$ is Lebesgue integrable on $(0,\infty )$. The sine power kernel integrability threshold is $1<p<2$: one endpoint requires $p<2$, and the other requires $p>1$. The signed improper integral for $f$ does converge by the Dirichlet test for improper integrals, but conditional improper convergence does not give a finite integral of $|f|$.