Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 18

Partition the finite interval as $a=x_0<x_1<\cdots <x_N=b$, with $\Delta x_i=x_{i+1}-x_i$. For a Riemann integrable function, the left-endpoint Riemann sums satisfyThe maximum subinterval width, rather than merely the number of subintervals, must tend to zero. Continuity of $f$ on the closed interval is a sufficient integrability hypothesis. More generally, any chosen sample point in each subinterval gives the same definite integral under these assumptions.

Index the stated geometric progression by $x_i=(1+ϵ{)}^i$, $i=0,1,\dots$. Then $\Delta x_i=ϵ(1+ϵ{)}^i$, and its left-endpoint contribution isHence the approximation is a geometric series withSince $ϵ>0$ and $\alpha >0$, the ratio satisfies $0<q<1$ and the series converges. The integrand is decreasing, so the left-endpoint sum is an upper bound for the improper integral.

The binomial series gives $(1+ϵ{)}^{-\alpha }=1-\alpha ϵ+O({ϵ}^2)$. ThereforeTo justify the limiting approximation over the infinite interval, also use the right-endpoint sum. It is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$, and monotonicity gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds have the same limit. The geometric partition bounds for a power-law improper integral and the squeeze theorem thus proveThis bounding argument matters because the geometric partition has unbounded widths far along the half-line; a finite-interval maximum-mesh argument cannot simply be applied to the entire infinite partition at fixed $ϵ$.

For $m=2$, evaluate the improper integral directly:For $m\ge 3$, use integration by parts with $u=x^{m-2}$ and $dv=(1+x{)}^{-(m+n-2)}dx$. Then $v=-(1+x{)}^{-(m+n-3)}/(m+n-3)$ andThe boundary term vanishes: at zero $m-2>0$, while at infinity its power is $1-n<0$. The restriction $m\ge 3$ belongs to this recurrence step; the separate $m=2$ base case must not be replaced by a formal zero times a divergent $I_{1,n}$.

Apply the recurrence repeatedly down to $m=2$:Including the base factor $1/(n-1)$ and rewriting the products as factorials givesThe formula also includes $m=2$ because $0!=1$. It is symmetric in $m,n$, consistent with the integration by substitution $x=1/t$. Equivalently it is the Euler beta function $B(m-1,n-1)$, although that identification is not needed for the recurrence derivation.