Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 20 / c

Write $I(x)=\ln g(x)$. For $x>-1$, differentiation under the integral sign givesIndeed, on any compact parameter interval $c\le x\le d$ with $c>-1$, the differentiated integrand is bounded by the integrable function $t^c$. The dominated convergence theorem therefore justifies the operation despite the improper lower endpoint. The apparent singularity at $t=1$ is removable.

Since $g=e^I$, the chain rule gives the first-order linear differential equation

The required range isNear $t=1$, the ratio $(t^x-1)/\ln t$ tends to $x$, so it causes no convergence problem. Near $t=0$, one useful identity isFor $x>-1$, this bounds its absolute value by $|x|t^{\min (x,0)}$, which is integrable on $(0,1)$. It also supplies the domination used in part (i).

For $x=-1$, the integrand is negative and behaves as $-1/[t|\ln t|]$ near zero; its improper integral diverges. For $x<-1$ its negative singularity is even stronger, and comparison with that same divergent integral again rules out convergence. Thus the original finite real-valued definition of $g$ has precisely the stated domain.

At $x=0$, the integrand vanishes for $0<t<1$, so $I(0)=0$ andSolving the first-order linear differential equation from part (i) on $(-1,\infty )$ gives $g=C(x+1)$. The initial condition fixes $C=1$, soEquivalently, the defining exponent is the logarithmic integral representation of the natural logarithm, $I(x)=\ln (x+1)$.

For $x>-1$, interpret the endpoint $-1$ by a one-sided limit. Part (iii) shows that $g$ extends continuously there with value zero, so both nested integrals converge. Let $F(x)$ denote the nested expression inside the derivative. The fundamental theorem of calculus, applied twice with moving lower endpoints, givesBy part (iii), $g^{\prime }(x)=1$, henceA direct check is ${\int }_s^{-1}(t+1)dt=-(s+1{)}^2/2$, so $F(x)=(x+1{)}^3/6$, with the same third derivative.