Past exam of the options II course of the University of Oxford / 2007 / D (Integration)

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open image, with continuously differentiable inverse. Equivalently one may specify a one-to-one continuously differentiable map whose Jacobian determinant is everywhere nonzero; the inverse function theorem gives the corresponding smooth inverse on its image.

The change-of-variables formula isFor an integrable $f$, the transformed function with its absolute-Jacobian factor is integrable as well. The absolute value is essential because the integral does not depend on whether the coordinate change preserves orientation.

Use $u=x-y$, $v=x+y$. It sends the diamond-shaped open square onto $(-1,1{)}^2$, and the inverse Jacobian has absolute determinant $1/2$. On this rectangle $|u|<1$ and $v+2>1$, so the original function is measurable, bounded by one and supported on a finite-area set, proving integrability. The transformed integral separates:The two one-dimensional integrals follow from the fundamental theorem of calculus; Fubini's theorem applies by the established integrability.

The dominated convergence theorem says that if measurable $f_n\to f$ almost everywhere and $|f_n|\le g$ for an integrable $g$, then $f$ is integrable, $\int |f_n-f|\to 0$, and consequently $\int f_n\to \int f$.

First derive Fatou's lemma from monotone convergence. For nonnegative $u_n$, the measurable functions $v_n={\inf }_{j\ge n}{u}_j$ increase to $\mathrm{liminf}{u}_n$, and $\int v_n\le {\inf }_{j\ge n}\int u_j$. Monotone convergence therefore yields $\int \mathrm{liminf}{u}_n\le \mathrm{liminf}\int u_n$.

Now $f$ is measurable and $|f|\le g$ almost everywhere, so is integrable. Apply Fatou's lemma to $u_n=2g-|f_n-f|\ge 0$, which converges almost everywhere to $2g$:The last integrals are nonnegative, so they tend to zero. Finally $|\int f_n-\int f|\le \int |f_n-f|\to 0$. This argument also works for complex-valued functions.

At each $x>0$, eventually $x<n$, and the stipulated power limit gives $f_n(x)\to e^{-x}\sin x$. At nonpositive $x$ the functions vanish. On their positive support,The supplied integrable exponential therefore dominates the sequence. Dominated convergence givesThe final improper integral is absolutely convergent, so agrees with its Lebesgue integral.