Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 20

For $\alpha \ne 0$, integration by substitution $t=\alpha x$ givesAt $\alpha =0$ the integrand is one, so $I(0)=1$, also the continuous limit of this expression. For nonzero $\alpha$, differentiation under the integral sign is valid on the finite interval and yieldsSet $\alpha =\sqrt{3}$ and use $\arctan \sqrt{3}=\pi /3$. The requested value is

The stated derivative formula needs a domain qualification. The original PDF calls $\alpha$ real, but the integral is even in $\alpha$; its derivative therefore cannot equal the printed positive expression on both sides of zero.

First take $\alpha >0$. The improper integral converges, with an $O(x^2)$ integrand near zero and $O(\ln x/x^2)$ decay at infinity. On compact positive parameter intervals, an integrable bound also justifies differentiation under the integral sign:For $\alpha \ne 1$, the partial fraction decomposition isIntegrating the rational functions givesAt $\alpha =1$, the same result follows by continuity, or directly from $2{\int }_0^{\infty }{x}^2/(1+x^2{)}^{2}dx=\pi /2$.

Integrating gives $E(\alpha )=\pi \ln (1+\alpha )+C$ on the positive half-line. To fix $C$, take $\alpha \to 0^{+}$: for $0\le \alpha \le 1$ the integrand is bounded by $\ln (1+x^2)/(1+x^2)$, an integrable function. The dominated convergence theorem gives $E(\alpha )\to E(0)=0$, so $C=0$. Evenness then extends the logarithmic parameter integral with Cauchy weight to its full stated domain:For negative $\alpha$, the derivative is $-\pi /(1-\alpha )$. At zero the right and left derivatives are $\pi$ and $-\pi$, so $E$ is not differentiable there. The pointwise derivative of the integrand at zero is zero, but interchanging it with the infinite integral at that parameter is invalid. Finally, substituting $\alpha =1$ gives the requested special value