Past exam of the options II course of the University of Oxford / 2009 / D / D3

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