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

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 $\arctan x$.

On $(0,1)$, the integrand is bounded by one. For $x\to \infty$, $\alpha x^{\beta }-2\log x\to \infty$, since every positive power grows faster than the natural logarithm. Thus $e^{-\alpha x^{\beta }}\le x^{-2}$ eventually. The comparison test for integrals proves Lebesgue integrability.

To prove joint continuity, fix $({\alpha }_0,{\beta }_0)$ with both coordinates positive. In a sufficiently small parameter rectangle, $\alpha \ge a:={\alpha }_0/2>0$ and $\beta \ge b:={\beta }_0/2>0$. For every parameter pair in this rectangle,The first argument shows that $H$ is integrable. Along any parameter sequence tending to $({\alpha }_0,{\beta }_0)$, the integrands converge pointwise, and the same $H$ dominates their eventual tail. The dominated convergence theorem gives $g(\alpha ,\beta )\to g({\alpha }_0,{\beta }_0)$. The sequential criterion for continuity proves the required joint continuity throughout the positive quadrant.

For comparison, the change of variables $u=\alpha x^{\beta }$ identifies the value with a gamma function:The domination argument establishes continuity directly, without needing a separate theorem about parameter dependence of the gamma function.

For each nonnegative integer $k$, the change of variables $x=e^{-t}$ givesThe gamma function identity $\Gamma (k+1)=k!$ follows by repeated integration by parts. Since $\log x<0$ on $(0,1)$, the sign of the original integrand is $(-1{)}^k$, soFor $k=0$, the integral is one, consistent with the formula.

The exponential series gives $x^x=e^{x\log x}={\sum }_{k=0}^{\infty }(x\log x{)}^k/k!$. Since $|x\log x|\le 1/e$ on $(0,1)$, the Weierstrass M-test proves uniform absolute convergence, using the summable bounds $e^{-k}/k!$. Therefore one can integrate the series term by term over this finite interval. Using part i,This is the alternating identity in sophomore's dream. The same estimate also proves absolute convergence of the integrated series; no cancellation is needed to justify the interchange.