Past exam of the integration course of the University of Oxford / 2021 / 2

Near zero, $\sin x/x\to 1$, so $|\sin x|/x^{\alpha }$ is comparable to $x^{1-\alpha }$. The power integral test shows that the integral near zero is finite exactly when $\alpha <2$.

For the tail, $|\sin x|\le 1$ proves integrability when $\alpha >1$. To prove necessity, take the disjoint intervalsOn each, $\sin x\ge 1/2$, and $x\le 2\pi (k+1)$. Therefore ${\int }_{I_k}|\sin x|x^{-\alpha }dx\ge c_{\alpha }(k+1{)}^{-\alpha }$ with $c_{\alpha }>0$. The p-series diverges when $\alpha \le 1$, proving that the absolute tail integral diverges in that range.

Combining the two conditions,For $0<\alpha \le 1$, the tail improper integral converges conditionally by the Dirichlet test for improper integrals, but that does not give Lebesgue integrability.

The absolute value is bounded by $e^{-ax}$, whose integral is $1/a$. Thus the function is Lebesgue integrable. The complex integral of an exponential function isTaking the real part, justified by absolute integrability, gives

Again $|e^{-ax}\sin (bx)|\le e^{-ax}$ proves Lebesgue integrability. Taking the imaginary part of the absolutely convergent complex integral in part i gives

The function is not Lebesgue integrable. Fix $0<\eta <\pi /2$ and restrict to $x\ge 1$, $0<y<\eta$. There $\cos y\ge \cos \eta >0$. Tonelli's theorem for the nonnegative absolute value gives the lower boundThe last integral diverges. Indeed, on $[k\pi ,k\pi +\pi /6]$, $|\cos 2x|\ge 1/2$; each such interval contributes at least a positive constant times $1/(k+1)$. The harmonic series diverges. Hence the integral of $|g|$ is infinite, regardless of possible signed cancellations. One cannot infer Lebesgue integrability by formally applying Fubini's theorem to the oscillatory integrand.

The function is Lebesgue integrable. The inequalities $|\sin u|\le \min \{u,1\}$ for $u>0$ give bounds that handle the two ends of the $x$-axis differently. For $0<x<1$,For $x\ge 1$,Tonelli's theorem therefore provesThe factors of sine suppress the near-axis contributions that caused divergence in part i.