Past exam of the options II course of the University of Oxford / 2006 / D (Integration)

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise almost everywhere and the functions are measurable, then $\int h_n\uparrow \int h$, allowing the value infinity.

Interpret the displayed densities as zero outside $(0,1)$, so no undefined logarithm is evaluated there. Each $f_k$ is nonnegative and measurable. With $x=e^{-t}$,where the last equality is integration by parts. This is finite, proving Lebesgue integrability.

The nonnegative partial sums satisfy, for $0<x<1$,They vanish outside the interval as well. Monotone convergence therefore givesThe series is finite by comparison with the convergent series ${\sum }_{j\ge 1}{j}^{-2}$, whose convergence follows from the integral test. Since $g\ge 0$, this finite integral proves absolute Lebesgue integrability as required.

We first prove the differentiation theorem. Fix $t\in (0,b)$ and take nonzero $h\to 0$ with $t+h\in (0,b)$. The measurable difference quotientsconverge pointwise to $f_t(x,t)$. The mean value theorem and the assumed derivative bound give $|q_h(x)|\le g(x)$. In particular $f_t(\cdot ,t)$ is measurable as a limit of quotients along a sequence, and its bound by $g$ proves integrability. The dominated convergence theorem, in its parameter-limit form, says that pointwise convergence with a common integrable bound permits convergence of the integrals. Applying it yieldsThe parameter-limit statement follows from the ordinary sequential theorem by applying it to every sequence of nonzero $h$ tending to zero.

For the specific function, $|f(x,t)|\le e^{-x^2}$, which is integrable by the allowed fact. It is differentiable in $t$, withThe last majorant has integral one on $[0,\infty )$, using the substitution $u=x^2$. All three hypotheses therefore hold. The indicator of the parameter domain is one at every point under consideration and does not affect this derivative.

The theorem givesIntegration by parts has zero boundary term: the sine vanishes at zero and the Gaussian tends to zero at infinity. Hence $F^{\prime }(t)=-2t{\int }_0^{\infty }{e}^{-x^2}\cos (2xt)dx=-2tF(t)$.

The same Gaussian majorant and dominated convergence give continuity at zero, where the allowed Gaussian integral gives $F(0)=\sqrt{\pi }/2$. The differential equation implies $(e^{t^2}F(t){)}^{\prime }=0$; continuity at zero fixes that constant. Thus