Past exam of the integration course of the University of Oxford / 2021 / 1 / a / Solution

Extend the integrand by zero to all of $(0,\infty )$:For each fixed $x>0$, eventually the indicator function is one and $\sin (x/n)/(x/n)\to 1$. Hence $h_n(x)\to (1+x^2{)}^{-1}$ pointwise. On its support, $x/n\le 2\pi$, so $|\sin t|\le t$ for $t\ge 0$ givesThe final bound is Lebesgue integrable on $(0,\infty )$ and independent of $n$. The dominated convergence theorem therefore givesThe last integral follows from the antiderivative $\arctan x$.