Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / D (Integration)

The function is nonnegative and measurable. For $\alpha \ne -1$, integrate from $\epsilon$ to one to get $(1-{\epsilon }^{\alpha +1})/(\alpha +1)$. The limit is finite exactly when $\alpha +1>0$. At $\alpha =-1$ the integral is $-\log \epsilon \to \infty$. Thus $\boxed{x^{\alpha }\in L^1(0,1)⟺\alpha >-1}$.

Integrating from one to $R$ gives $(R^{\alpha +1}-1)/(\alpha +1)$ for $\alpha \ne -1$. It has a finite limit as $R\to \infty$ exactly when $\alpha +1<0$, while at $\alpha =-1$ the logarithm diverges. Nonnegativity equates finiteness of this integral with Lebesgue integrability. Hence $\boxed{x^{\alpha }\in L^1(1,\infty )⟺\alpha <-1}$.

On $(0,1)$ its absolute value is at most $x|\log x|$, which has integral $1/4$, by integration by parts or the substitution $x=e^{-t}$. The function is measurable and this majorant is integrable, so $\boxed{x\log x/(1+x{)}^2\text{ is integrable on }(0,1)}$.

For $x\ge 1$, $(1+x{)}^2\le 4x^2$, so $x\log x/(1+x{)}^2\ge (\log x)/(4x)\ge 0$. The latter integral diverges because its primitive is $(\log x{)}^2/8$. Thus the absolute integral diverges and the answer is $\boxed{\text{not integrable on }(1,\infty )}$.

The integrand is nonnegative and measurable. It is bounded by $x$ near zero and by $x{e}^{-x}$ for $x>0$, so its integral is finite. For every $x>0$ the geometric expansion gives $x/(e^x+1)={\sum }_{n\ge 1}(-1{)}^{n-1}x{e}^{-nx}$. To justify termwise integration, note thatTonelli gives an integrable sum of absolute values; dominated convergence then applies to the signed partial sums. ConsequentlyIf evaluated further, this is $(1-2^{-1})\zeta (2)={\pi }^2/12$.

For product measure spaces, Tonelli's theorem states that for a nonnegative measurable function the product-space integral equals either iterated integral, allowing the value $+\infty$. Fubini's theorem states that if a measurable function has finite integral of its absolute value, then its sections are integrable almost everywhere, both iterated integrals exist and equal the product-space integral. The absolute-integrability hypothesis is what permits this conclusion for a signed function. These statements apply to the Lebesgue product spaces used here, which are sigma-finite.

The function $g$ is measurable, and for fixed positive $y$ both terms are integrable over $0<x<1$ by linear substitution and integrability of $f$. Those substitutions giveThe strict sign follows from positivity of $f$ on a nonempty interval, since $a>1$. Similarly each section in $y$ is absolutely integrable for fixed $x>0$, andIf $g$ were absolutely integrable on the stated rectangle, Fubini would make the two finite iterated integrals equal. The first would be strictly negative and the second strictly positive, which is impossible. Therefore $\boxed{g\notin L^1((0,1)\times (1,\infty ))}$. The one-dimensional section integrability does not rescue the missing product-space hypothesis.

The function $h$ is measurable and nonnegative. In first-quadrant polar coordinates, including the Jacobian, its integral isThe radial factor is $1/2$ by the supplied fact and the angular integrand is at most one, giving finiteness. The nonnegative change-of-variables formula is valid before finiteness is known, and this bound proves $h\in L^1$.

Fubini may now factor the Cartesian integral. Substituting $u=x^2$, $v=y^2$ in the respective factors givesBoth are finite: their powers at zero exceed $-1$, and the exponential controls infinity. Comparing with the polar expression proveswhich is exactly the product of integrals requested.