Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 2 / a / Solution

Let $F(a)$ be the given integral. Its integrand extends continuously to $x=0$, with value $a$. The Leibniz integral rule includes both the moving upper endpoint and the derivative of the integrand. For $a\ne 0$ it givesConsequentlyThe zero case follows directly from $F(a)=a^3+O(a^9)$ near zero: expand $\sin (ax)/x=a+O(a^3{x}^2)$ and integrate up to $a^2$. Equivalently, the change of variables $t=ax$ gives $F(a)={\int }_0^{a^3}(\sin t)/tdt$, whose derivative is the same expression, interpreted continuously at zero. This substitution works for negative $a$ as an oriented integral too.