Past exam of the options II course of the University of Oxford / 2012 / D (Integration)

The monotone convergence theorem states that if measurable functions $0\le f_n\uparrow f$ pointwise on a measure space, then $\int f_n\uparrow \int f$, allowing the limit to be infinite.

Extend each restriction to $I_n$ by zero on $I\setminus I_n$. The functions $f{1}_{I_n}$ are nonnegative and increase pointwise to $f$ because the intervals are increasing and exhaust $I$. HenceFor nonnegative $f$, integrability means finiteness of this integral, soNo boundedness or closedness of the intervals is required.

The function is nonnegative, measurable, and even. Direct integration givesThus it is integrable, with integral $\boxed{2}$.

On finite intervals the function is continuous, butConsequently its nonnegative integral on the half-line is infinite: it is not integrable. The divergence occurs at infinity, not at the origin.

For every real $x$, the numerator is at most $2e^{|x|}$, while the denominator is at least $e^{2|x|}$. ThereforeThe upper bound is integrable by part (a), so the comparison theorem shows that the function is integrable on $\mathbb{R}$.

We prove absolute integrability. On $(0,1]$, $|\sin (1/x)|\le 1$ and the denominator is at least one, so $|f(x)|\le 1$. On $[1,\infty )$, the standard inequality $|\sin t|\le |t|$ givesSubstitution $t=\log x$ yieldsThus ${\int }_0^{\infty }|f(x)|dx\le 1+\pi /2<\infty$, and the function is integrable despite its oscillation near zero.

For a continuously differentiable diffeomorphism $\varphi :J\to I$ between intervals, the integration by substitution statesIt holds as an identity element of extended integrals for nonnegative measurable $h$. For signed or complex-valued measurable $h$, integrability on one side is equivalent to integrability after substitution, and the identity element then holds. The absolute derivative accounts for a decreasing substitution; on oriented finite intervals the equivalent formula uses ${\varphi }^{\prime }$ and oriented endpoints.

Put $C={\int }_0^{\infty }(1+x^4{)}^{-1}dx=\pi /(2\sqrt{2})$. The map $x=y^{-1/4}$ is a decreasing diffeomorphism of the positive half-line onto itself andApplying the nonnegative version before assuming finiteness proves both existence and value:Tonelli's theorem permits either iterated integral of a nonnegative measurable function on a product of sigma-finite measure spaces and identifies both with its extended product integral. Fubini's theorem states that for an absolutely integrable measurable function on such a product, almost every section is integrable, the section-integral functions are integrable, and the two iterated integrals equal the product integral.

The given two-variable function is continuous and positive on the positive quadrant. For fixed $y>0$, substituting $t=x{y}^{1/4}$ givesTonelli now givesThis proves absolute integrability, so Fubini is applicable.

For $x\ne 1$, integrate the displayed partial fraction identity element in $y$ up to $R$:At $x=1$ the section integral is ${\int }_0^{\infty }(1+y{)}^{-2}dy=1$, agreeing with the continuous limit of this expression. Therefore Fubini identifies the remaining integral and givesThe apparent singularity at one is removable, with limiting integrand value $1/4$. Positivity and the finite product integral already establish convergence at both ends.