Past exam of the integration course of the University of Oxford / 2015 / 2

The function is continuous for $x>0$, hence Lebesgue measurable. Near zero, the bound $|\sin x|\le x$ givesThus $f$ is Lebesgue integrable on $(0,1]$ for $q<2$. The convergence is an instance of power singularity integrability.

For the tail, take the pairwise disjoint intervalsOn each $I_k$, $|\sin x|\ge 1/2$, $x\le (k+1)\pi$, and the interval length is $2\pi /3$. ThereforeSumming the nonnegative integrals over these disjoint intervals yieldsby divergence of the p-series. Hence $f$ is not Lebesgue integrable on $[1,\infty )$ for $0<q\le 1$.

The test is absolute integrability: oscillatory sign cancellation does not make the absolute-value integral finite. A conditionally convergent improper integral is a different notion. The sine power kernel integrability criterion separates these endpoint thresholds.

DefineThis is a Lebesgue integrable function: on $(0,1]$ it is bounded by $x^{-1/2}$, whose integral is two, while on $[1,\infty )$ it is bounded by $e^{-x}$, whose integral is $e^{-1}$. Since $|\sin (xy)|\le x|y|$,Thus $h_y$ is Lebesgue integrable for every $y\ge 0$, including $h_0=0$; the same argument actually works for all real $y$.

The mean value theorem for the sine gives $|\sin u-\sin v|\le |u-v|$. Henceand the triangle inequality for an integral givesThus $F$ is a globally Lipschitz function, in particular continuous on $[0,\infty )$, including the endpoint zero.

For the derivative, fix $y>0$ and take nonzero $t$ sufficiently small that $y+t>0$. The difference quotient has the integrandFor every $x>0$, $D_t(x)\to H(x)\cos (xy)$ as $t\to 0$. The same mean value theorem gives the uniform-in-$t$ bound $|D_t(x)|\le H(x)$.

The dominated convergence theorem states that if measurable functions $u_n$ converge almost everywhere to $u$ and $|u_n|\le G$ for a single Lebesgue integrable function $G$, then $u$ is Lebesgue integrable and $\int |u_n-u|\to 0$, hence $\int u_n\to \int u$. Apply it to the difference quotients along every sequence $t_n\to 0$, using the common dominator $H$. The resulting limit is independent of the sequence, soThe right-hand integral is absolutely convergent because its integrand has absolute value at most $H$. This is dominated differentiation under the integral sign, justified by an explicit integrable bound rather than a formal interchange.

For every fixed $x>0$, $\log (x+n)/n\to 0$, so $f_n(x)\to 0$. To exchange the limit with the Lebesgue integral, we need a common integrable bound over the whole half-line.

For $n\ge 1$, $x+n\le n(1+x)$, soTogether with $|\sin x|\le \min (x,1)$, this givesFor $0<x\le 1$, $\log (1+x)\le x\le 1$, hence $G(x)\le 2x^{1-q}$, which is integrable because $q<2$. For $x\ge 1$,Both terms are integrable because $q>1$:The second equality follows from integration by parts, or the substitution $x=e^t$. Consequently $G$ is a Lebesgue integrable function, and every $f_n$ is Lebesgue integrable.

The dominated convergence theorem states that almost everywhere convergence of measurable functions under one integrable absolute-value bound implies L1 convergence, and in particular convergence of their integrals. Here the pointwise limit is zero and the dominator is $G$, soIn fact we obtain the stronger conclusion $\Vert f_n{\Vert }_1\to 0$. A quantitative version follows by retaining the factors $1/n$ in the bound:The preceding endpoint estimates show $A,B<\infty$, so the L1 norm is $O(\log n/n)$ as $n\to \infty$, with $O$ denoting Big O notation. This is logarithmic reweighting under dominated convergence.