Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 20

Under the usual regularity assumptions, the Leibniz integral rule isThere is a genuine domain problem in the following integral as printed in the PDF. PutThe point $t=2$ is a simple pole, withFor every $x>4$, the lower endpoint $\sin (1/x)$ is less than $2$ and the upper endpoint $\sqrt{x}$ is greater than $2$. The interval therefore crosses the pole. Since $q(t)=33/[7(t-2)]+O(1)$ near $2$, the two one-sided improper integrals diverge logarithmically, with opposite signs. Their divergences do not define an ordinary integral. Consequently the ordinary limit in the question is undefined as written: its underlying integral has no finite value for any $x\ge 4$.

A separately specified Cauchy principal value does exist for $x>4$. Under that interpretation, differentiating a principal-value integral across a fixed simple pole givesThe first term isand the second tends to zero because $q$ is bounded near $t=0$. Thus the principal-value version has limit $\boxed{1}$. The two conclusions are distinct: this regularised answer requires the additional principal-value convention, which the printed question does not state.

First suppose $\alpha >0$. Differentiation under the integral sign, justified on compact positive parameter intervals by an integrable bound, givesThe first integral is $f(\alpha )/\alpha$. For $\alpha \ne 1$, use the partial fraction decompositionThe standard inverse tangent antiderivatives then giveThe same value holds at $\alpha =1$ by continuity, or by direct integration. Hence the proposed first-order linear differential equation is valid on the positive parameter range:Divide by $\alpha$ to recognise a quotient rule derivative:To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead considerThis limit follows from the dominated convergence theorem: for $0<\alpha \le 1$ and $0<x<1$, the logarithm's absolute value is bounded by $2|\ln x|+\ln 2$; for $x\ge 1$, it is at most $\ln (1+x^2)$. After division by $1+x^2$, these are integrable. The limiting improper integral is absolutely convergent, and the substitution rule $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. ThusThe question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining integral is odd in $\alpha$, because its logarithm depends on ${\alpha }^2$. Applying the positive-parameter result to $|\alpha |$ therefore gives the correct extensionwith $f(0)=0$. In particular, for $\alpha <0$, the right-hand side of the differential equation must be $-\pi \alpha /(1-\alpha )$; for every $\alpha \ne 0$ the unified expression is $\pi |\alpha |/(1+|\alpha |)$. The logarithmic quadratic parameter integral therefore provides the requested identity on its valid range and the corrected answer elsewhere.