Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1 / 11D / Solution

For a bounded real function on $[a,\infty )$, convergence of its improper integral means that its restriction to every $[a,b]$, $b>a$, is Riemann integrable and that the finite limitexists. Boundedness by itself does not guarantee either local Riemann integrability or existence of this limit.

A monotone function on $[a,b]$ is bounded between its endpoint values. Divide the interval into $m$ equal pieces with endpoints $x_j=a+j(b-a)/m$. If $f$ is nondecreasing, its upper and lower Darboux sums differ byFor nonincreasing $f$, the same calculation with reversed endpoint values gives $(b-a)(f(a)-f(b))/m$. Thus in either case $U-L=(b-a)|f(b)-f(a)|/m\to 0$. For every $\epsilon >0$ one can choose $m$ to make this gap smaller than $\epsilon$; the Riemann integrability criterion proves that $f$ is Riemann integrable.

Now let $f$ be decreasing on $[1,\infty )$ with limit zero. For each $x$, $f(x)\ge {\lim }_{y\to \infty }f(y)=0$. On $[n,n+1]$ its monotonicity givesSumming these bounds, with $S_N={\sum }_{n=1}^{N}f(n)$, yieldsBoth the partial sums $S_N$ and the truncated integrals are nondecreasing because $f\ge 0$. The inequalities show that one family is bounded above if and only if the other is. The bounded monotone sequence theorem then gives convergence of the sums exactly when the integer-endpoint integrals converge. For real $b$ between $N$ and $N+1$, the integral to $b$ lies between the integrals to these integer endpoints; the squeeze theorem supplies the same limit. If either family is unbounded, both tend to $+\infty$. This proves the integral test for convergence in the required form.

For $\alpha <0$, use $f(x)=x^{\alpha }$, which is decreasing and tends to zero. Its truncated improper integral isThis has a finite limit exactly when $\alpha <-1$. If $\alpha \ge 0$, the terms $n^{\alpha }$ do not tend to zero, so the term test for divergence applies. Thus the full p-series criterion isFor every $\alpha \ge -1$ its nonnegative partial sums tend to $+\infty$.