Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 17

The fundamental theorem of calculus, using continuity of $f$ near $x$, givesThe lower endpoint $a$ is constant, so it contributes no additional term.

On the negative half-line, differentiation gives $d\ln (-u)/du=1/u$. The integration path from $-1$ to $x<0$ stays away from the singularity at zero. ThereforeThis choice satisfies $F(-1)=0$.

A suitable real antiderivative on the disconnected domain isFor $x>0$, $G^{\prime }=1/x$ directly; for $x<0$, $G=\ln (-x)$ and the chain rule again gives $G^{\prime }=1/x$. The graph is even, crosses zero at $x=\pm 1$, tends to $-\infty$ on either side of zero, and tends to $+\infty$ as $|x|\to \infty$. Independent additive constants may be chosen on the two disconnected half-lines; both are set to zero here.

At $u=1$, the integrand is unbounded but integrable: $1/\sqrt{u^2-1}\sim [2(u-1){]}^{-1/2}$. Thus the lower-limit integral is understood as an improper integral. With $z=u+\sqrt{u^2-1}$,The lower endpoint corresponds to $z=1$, givingThe limiting value at $x\downarrow 1$ is zero. Here $\mathrm{arcosh}$ is the inverse hyperbolic cosine.

A single formula valid on both real components isLet $s=\sqrt{x^2-1}>0$. Since $x+s\ne 0$,Equivalently, $R(x)=\mathrm{arcosh}x$ for $x>1$ and $R(x)=-\mathrm{arcosh}(-x)$ for $x<-1$. The graph is odd for this choice of constants, increases on both components, and approaches zero with vertical tangent as $x\to 1^{+}$ or $x\to -1^{-}$. Its far-field behaviors are $R(x)\sim \ln (2x)$ on the right and $R(x)\sim -\ln (-2x)$ on the left. The interval $[-1,1]$ is excluded. Independent additive constants could again be assigned to the two components. The right panel sketches these real inverse-hyperbolic antiderivative branches.

Let $H(u)={\sin }^{2}u+\ln u$. The chain rule gives $H^{\prime }(u)=\sin 2u+1/u$, so the two given summands combine into its logarithmic derivative, $f(u)=H^{\prime }(u)/H(u)$. For $u>1$, $H(u)>0$, so the real logarithm is defined throughout the integration path. Since $H(\pi )=\ln \pi$,This satisfies $F(\pi )=0$, and differentiation recovers both terms of the integrand.