Past exam of the options II course of the University of Oxford / 2010 / D (Integration)

The monotone convergence theorem states that for nonnegative measurable functions $f_n\uparrow f$ almost everywhere, $\int f={\lim }_n\int f_n$, allowing the value $+\infty$.

The dominated convergence theorem states that if measurable functions $f_n\to f$ almost everywhere and $|f_n|\le g$ almost everywhere for some integrable $g$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$ as well. Both statements apply on any measure space.

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. Multiplying by $(\log x{)}^2$ gives nonnegative terms, so the monotone convergence theorem permits termwise integration. Put $x=e^{-t}$ to obtainwhere scaling followed by two integrations by parts evaluates the last integral. HenceThe finite value also establishes integrability at both endpoints.

A sufficient differentiation under the integral sign theorem is the following local domination criterion. Suppose $u(t,\cdot )$ is measurable for each $t$, $u(t_0,\cdot )$ is integrable at one point of a neighborhood $J$ of $t_0$, and for almost every $x$ the function $u(\cdot ,x)$ is differentiable on $J$. Assume an integrable function $g_J$ satisfies $|{\partial }_{t}u(t,x)|\le g_J(x)$ for $t\in J$ and almost every $x$. Then $I$ is differentiable on the interior of $J$ andThe mean value theorem bounds the difference quotients by $g_J$, and the dominated convergence theorem passes their pointwise limit through the integral. Applying this criterion locally throughout $(a,b)$ proves differentiability there. One may impose the stronger condition that every $u(t,\cdot )$ is integrable from the outset.

For every integer $j\ge 0$, the $j$th derivative with respect to $t$ of $\cos (t\cos x)$ is $(\cos x{)}^j$ times a sine or cosine with a sign. Its absolute value is at most $1$ on the finite interval $[0,\pi /2]$. This is an integrable dominating function independent of $t$. Repeated differentiation under the integral sign is therefore justified for every order, giving $J_0\in C^{\infty }(\mathbb{R})$.

Differentiating the integral representation givesNow integrate the derivativeThe boundary terms vanish at both endpoints. Consequently $J_0^{\prime }(t)=-t(J_0^{\prime \prime }(t)+J_0(t))$, provingThe identity includes $t=0$, where it reads $J_0^{\prime }(0)=0$.

A naive absolute bound on the original derivative would be $1/x$, which is not integrable on $[1,\infty )$. Instead make the change of variable $s=tx$, valid for $t>0$:The integral is absolutely convergent for every positive lower endpoint. Its integrand is continuous there, so the fundamental theorem of calculus givesThe product rule now yieldsThis proves differentiability and the equation without an unjustified differentiation of a conditionally convergent derivative integral.