Past exam of the integration course of the University of Oxford / 2014 (A4, 4323)

Source release: Oxford examination archive, member FOI-20210708-1 - Data/4323.pdf. The cover dates the examination to 19 June 2014 and prints formal paper code 4323.

Answer availability checked on 28 October 2026: a bounded search of official Oxford sources found no advertisement covering complete or selected-part answers for this exact 2014 paper. The department's general policy describes even-year solutions dating back to at least 2016; that statement does not identify coverage of this earlier paper.^[ref]

The monotone convergence theorem states: on a measure space $(X,\mathcal{F},\mu )$, let $u_n:X\to [0,\infty ]$ be measurable functions withThen $u$ is measurable andThe integrals and their limit may be infinite. No integrable dominator, finite measure hypothesis, or sigma-finiteness is required. The nonnegativity and increasing order are the essential hypotheses in this form of the theorem.

Let $S_N(x)={\sum }_{n=1}^N|f_n(x)|$. These are nonnegative measurable functions increasing to the extended-valued function $S(x)={\sum }_{n=1}^{\infty }|f_n(x)|$. The monotone convergence theorem givesTherefore $S$ is finite almost everywhere: if $S=\infty$ on a measurable set of positive Lebesgue measure, its integral would exceed every finite bound. Hence the series ${\sum }_n{f}_n(x)$ converges absolutely outside the null set $N=\{S=\infty \}$. This is summable L1 norms imply almost-everywhere absolute convergence.

Define $s(x)={\sum }_n{f}_n(x)$ off $N$ and set $s=0$ on $N$. The function is Lebesgue measurable, being the pointwise limit of measurable partial sums off a measurable null set. Since $|s|\le S$ almost everywhere,so the sum is Lebesgue integrable.

For $s_N={\sum }_{n=1}^N{f}_n$, the absolute tail satisfiesApply the monotone convergence theorem to the nonnegative tail partial sums. We obtainThus the partial sums converge in L1 norm. The triangle inequality for an integral yieldsConsequentlyThe scalar series on the right also converges absolutely, since ${\sum }_n|{\int }_E{f}_n|\le {\sum }_n\Vert f_n{\Vert }_1<\infty$. This proves integration of a series with summable L1 norms without needing pointwise convergence on the exceptional null set.

The function is continuous for $x>0$, hence Lebesgue measurable. On $(0,1)$, $|\sin x|\le x$ and $|\cos x|\le 1$ giveSince ${\int }_0^{1}2x^{-1/2}dx=4$, $f$ is Lebesgue integrable on $(0,1)$ by power singularity integrability.

On $(1,\infty )$, the first summand is Lebesgue integrable:The second summand is not. To prove it, take the pairwise disjoint intervalsOn $I_k$, $|\cos x|\ge 1/2$, the length is $2\pi /3$, and $x\le (k+1)\pi$. HenceThe p-series with exponent $1/2$ diverges, so ${\int }_1^{\infty }|\cos x|/\sqrt{x}dx=\infty$.

If $f$ were Lebesgue integrable on this tail, subtracting its integrable first summand would make $\cos x/\sqrt{x}$ integrable as well, a contradiction. ThereforeThis uses the general fact that an integrable perturbation cannot repair nonintegrability. The conclusion concerns absolute integrability; oscillatory cancellation in an improper integral is a separate issue.

The dominated convergence theorem states: on a measure space $(X,\mathcal{F},\mu )$, let $u_n$ be measurable functions converging almost everywhere to a measurable $u$. Suppose there is a nonnegative measurable function $G$ with ${\int }_{X}Gd\mu <\infty$ and $|u_n|\le G$ almost everywhere for every $n$. Then $u$ is integrable andHere integrability means integrability with respect to the given measure; for Lebesgue measure it is Lebesgue integrability. Neither a finite measure space nor monotonicity of the sequence is required. The same single integrable dominator must control the whole sequence.

Fix $t_0\ge 0$. On a compact parameter interval containing all nonnegative $t$ sufficiently close to $t_0$, the corresponding rectangle with $x\in [0,1]$ is compact. The continuous function $f$ is therefore bounded there by some $M<\infty$, by the extreme value theorem.

For any sequence $t_n\ge 0$ tending to $t_0$, all sufficiently late $t_n$ lie in this rectangle's parameter range. For each $x\in [0,1]$, continuity givesThe constant function $M$ is integrable on $[0,1]$. The dominated convergence theorem therefore givesThe sequential criterion for continuity on a real interval proves $F$ is continuous on $[0,\infty )$. At zero this is right-continuity. This is continuity of parameter integrals on compact rectangles; the bound is local in the parameter, so $f$ need not be bounded on the entire unbounded strip.

We prove the required differentiation under the integral sign directly. Fix $t_0>0$, and take a compact parameter interval $J\subseteq [0,\infty )$ containing $t_0$ in its interior. The jointly continuous partial derivative ${\partial }_{t}f$ is bounded on the compact rectangle $J\times [0,1]$ by some $M_1$.

For nonzero $h$ sufficiently close to zero, defineFor each $x$, the mean value theorem in the parameter gives $|D_h(x)|\le M_1$. The definition of the partial derivative also gives $D_h(x)\to {\partial }_{t}f(t_0,x)$. Since $M_1$ is integrable on $[0,1]$, dominated convergence applied along any sequence $h_n\to 0$ yieldsThusAt $t=0$, repeat the argument on $J=[0,\delta ]$ with $h>0$. It proves that the right derivative exists and satisfies the same formula at zero. This is the appropriate meaning of differentiability on the closed half-line. In fact $F^{\prime }$ is continuous there: apply part (i) to the continuous function ${\partial }_{t}f$.

The proof establishes differentiation of parameter integrals on compact rectangles under the exact regularity assumptions used here. In particular, it supplies the theorem's proof instead of assuming an interchange of a limit and an integral.

Remove the apparent singularity at $x=0$ by definingThis is jointly continuous: write $\varphi (t,x)=t\mathrm{sinc}(tx)$ using the unnormalized sinc function, whose value at zero is one. Its partial derivative in $t$ isalso at $x=0$, since there $\varphi (t,0)=t$. This partial derivative is jointly continuous on the whole strip.

The value at the single endpoint $x=0$ does not change a Lebesgue integral. Part (b), whose dominated convergence proof applies to $\varphi$, therefore makes $g$ continuously differentiable on $[0,\infty )$, with a right derivative at zero, andFor $t>0$, direct integration givesAt zero the same integral gives $\boxed{g_{+}^{\prime }(0)=1}$, so no division by zero is required. Also $g(0)=0$. The substitution rule $u=tx$ identifies $g(t)={\int }_0^t\sin u/udu$ for $t>0$, the sine integral.

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be measure spaces with sigma-finite measures, and use their product measure.

Fubini's theorem: if $u$ is a measurable absolutely integrable function on $X\times Y$, then its sections are integrable for almost every fixed value of the other coordinate, the resulting section-integral functions are integrable, andAll three integrals are finite. Values of an inner integral on an exceptional null set can be assigned arbitrarily.

Tonelli's theorem: if $u:X\times Y\to [0,\infty ]$ is measurable and nonnegative, the same equality holds for its nonnegative product and iterated integrals, allowing the common value $+\infty$. No advance absolute integrability assumption is needed. On a completed Lebesgue measure product, the section and measurability assertions are interpreted almost everywhere.

Thus Tonelli's theorem can first establish that $\int |u|<\infty$; after this, Fubini's theorem permits signed integration in either order.

Assign any value, for example zero, at $(0,0)$, where the displayed quotient is undefined. A single point is a Lebesgue null set, so this choice does not affect Lebesgue integrability. Elsewhere the function is continuous, hence it is Lebesgue measurable on the plane after this extension.

In polar coordinates, $x^2+y^2=r^2$ and the area element is $rdrd\theta$. Applying the nonnegative change of variables formula to the absolute value givesFor $0<r\le 1$, $|\sin (r^2)|\le r^2$, so the radial integrand is bounded by $r^{3-2\alpha }$. This is integrable at zero when $\alpha <2$. For $r\ge 1$, $|\sin (r^2)|\le 1$, giving the integrable tail bound $r^{1-2\alpha }$ when $\alpha >1$. In particular,ThereforeEquivalently, the substitution rule $u=r^2$ reduces its exact L1 norm toThis is radial sine-power kernel integrability, reducing a two-dimensional question to the sine power kernel integrability thresholds at zero and infinity.

Fix $0<x<1$. On the open interval $(0,x)$, the function $y\mapsto (x-y{)}^{\beta -1}$ is finite, positive and continuous, and $f$ is nonnegative and Lebesgue measurable. Their product is therefore a nonnegative measurable function. Its Lebesgue integral is always defined as an element of $[0,\infty ]$, with no possibility of cancellation of two infinities. Thus $g(x)$ is well defined as a nonnegative extended integral.

The endpoint $y=x$ need not receive its formal power value: for $\beta <1$ that value would be singular. The integral is over $(0,x)$, and any chosen value at the endpoints has no effect because endpoints form a null set.

Finiteness is not implied by the present assumptions. For example, $f(y)=1/y$ is an admissible nonnegative measurable function. For any fixed $x$, the factor $(x-y{)}^{\beta -1}$ is bounded below by a positive constant on $0<y<x/2$, so $g(x)\ge C_x{\int }_0^{x/2}dy/y=\infty$. The later L1 norm bound will establish finiteness almost everywhere when $f$ is integrable. The construction is a Volterra power-kernel integral.

Define the joint function by an unambiguous piecewise prescription:It is nonnegative and Lebesgue measurable: on the open triangle the power factor is continuous and $f(y)$ is measurable for the product sigma-algebra, and the zero extension is measurable. This also resolves the diagonal singularity and avoids evaluating a noninteger power of a negative number outside the triangle. The source's inclusion of $y=x$ in its indicator set changes only a null set; set the value there to zero.

The function $g(x)={\int }_0^{1}H(x,y)dy$ is measurable by Tonelli's theorem. Apply that theorem to exchange the nonnegative integrals, with infinity allowed until the estimate proves otherwise:The inner integral is $(1-y{)}^{\beta }/\beta$ because $\beta >0$. The weight is at most one because $0<y<1$. Since $g\ge 0$, this provesIn particular $g$ is finite almost everywhere; any exceptional infinite values may be replaced by zero to obtain an everywhere finite representative. This is the L1 bound for a Volterra power-kernel integral.

An everywhere-finite conclusion would be stronger and can fail when $0<\beta <1$, even for integrable $f$. For example, put $x_0=1/2$ and $\gamma =(1+\beta )/2$, and take $f(y)=(x_0-y{)}^{-\gamma }$ for $0<y<x_0$ and zero otherwise. Since $\gamma <1$, $f$ is integrable. Butbecause $\beta -1-\gamma =(\beta -3)/2<-1$. This does not contradict Lebesgue integrability, which requires finiteness only almost everywhere.