Past exam of the options II course of the University of Oxford / 2009 / D (Integration)

The function is nonnegative and measurable, so Lebesgue integrability is equivalent to finiteness of its integral. For $\alpha \ne -1$,Letting $\epsilon \downarrow 0$ gives a finite limit exactly when $\alpha >-1$; at $\alpha =-1$ the integral is $-\log \epsilon$ and diverges. By the monotone convergence theorem for truncations,

For $\alpha \ne -1$,The limit as $R\to \infty$ is finite exactly when $\alpha <-1$; the case $\alpha =-1$ diverges logarithmically. Positivity and the monotone convergence theorem identify these limits with the Lebesgue integral. Hence

The function is measurable and negative on this interval. Its absolute value obeysThe final integral follows by integration by parts and $x^2\log x\to 0$ at zero. The comparison test for integrals proves integrability.

For $x\ge 1$, $(1+x{)}^2\le 4x^2$, soThe integral of the lower bound from one to $R$ is $(\log R{)}^2/8$, which diverges. Thus the nonnegative function has infinite integral and is not integrable.

The function is continuous and nonnegative, bounded above by $x$ on $(0,1)$ and by $x{e}^{-x}$ on $(1,\infty )$, so it is integrable. For $x>0$ the geometric series givesThe series is absolutely summable in integral becauseEach integral follows by substituting $t=nx$ and integrating $t{e}^{-t}$ by parts. Tonelli's theorem applied to the absolute terms, followed by Fubini's theorem, therefore justifies termwise integration and gives

On a product of sigma-finite measure spaces, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, allowing $+\infty$.

Fubini's theorem says that if a measurable function is absolutely integrable on the product, almost every section is integrable, the integrals of sections are integrable, and both iterated integrals equal the product integral. In particular, to establish absolute integrability one can first apply Tonelli's theorem to the absolute value and show its iterated integral is finite.

Both terms of each one-dimensional section are integrable by a positive scaling substitution, since $f\in L^1(0,\infty )$. For fixed $y>0$,For fixed $x>0$,The strict inequalities hold because $a>1$ and the measurable integrand is strictly positive on each nonempty interval. If $g$ were integrable on the stated product domain, Fubini's theorem would make both iterated integrals finite and equal. But the first order of integration would give a strictly negative value, and the other a strictly positive value. This contradiction proves $g$ is not integrable on that product.

The function is nonnegative and continuous on the positive quadrant. In polar coordinates its product integral is, by the nonnegative change of variables formula and Tonelli's theorem,The angular factor is at most one, while the supplied radial integral is $1/2$, so the product integral is at most $\pi /4$ and is finite. This establishes integrability before any signed interchange is used.

Let $A$ denote the angular integral. The polar evaluation is $A/2$. In Cartesian coordinates Fubini's theorem instead factors the integral asusing $u=x^2$, $v=y^2$. All factors converge: their powers exceed $-1$ at zero and the exponential controls infinity. Equating the evaluations gives