Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 20

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

First check the improper integral exists: near zero its numerator is $(\beta -\alpha )x^2+O(x^4)$, so the quotient is $O(x)$; at infinity both exponentials decay. For positive parameters in compact neighborhoods, the differentiated integrands have integrable bounds, justifying differentiation under the integral sign. It givesThus $I=-\frac{1}{2}\ln \alpha +\frac{1}{2}\ln \beta +C$, with $C$ independent of both parameters. Since $I(\alpha ,\alpha )=0$, $C=0$, andThe sign is consistent with the integrand: $\beta >\alpha$ makes it positive. Neither separate integral ${\int }_0^{\infty }{e}^{-\alpha x^2}dx/x$ converges at zero; it is the difference that is meaningful. The substitution $u=x^2$ identifies the result as a Frullani integral with an extra factor $1/2$.