Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / D / D3

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 $(1-2^{-1})\zeta (2)={\pi }^2/12$.