Prove that if f is monotone, nonnegative and unbounded on [a, ∞), then the improper integral of f from a to infinity necessarily diverges.

Step-by-step solution

1. 1

   Pin down what monotone + unbounded forces. A monotone function on $[a,\infty)$ is either non-decreasing or non-increasing. A non-increasing function that starts at $f(a)$ and stays $\ge 0$ is trapped in $[0,f(a)]$ — hence bounded. So the hypothesis of unboundedness rules that out: $f$ must be non-decreasing, and $f(x)\to+\infty$.

2. 2

   Extract a convenient threshold. Since $f(x)\to+\infty$, there is some $b\ge a$ with

   $f(b)\ge 1$

   and because $f$ is non-decreasing, $f(x)\ge 1$ for all $x\ge b$. The choice of the constant $1$ is arbitrary — any positive number works.

3. 3

   Compare on the tail. For any $T>b$, monotonicity of the integral gives

   $\int_{b}^{T}f(x)\,dx\ge \int_{b}^{T}1\,dx=T-b$

4. 4

   Let T grow. The right-hand side $T-b\to+\infty$, so $\int_{b}^{T}f\to+\infty$. Adding the finite piece $\int_{a}^{b}f$ (finite because $f$ is monotone, hence integrable, on the bounded interval $[a,b]$) does not change that:

   $\int_{a}^{T}f(x)\,dx\to+\infty\quad\text{as }T\to+\infty$

   $\boxed{\text{The improper integral diverges}}$

5. 5

   Check the necessity of each hypothesis. Drop unbounded and the conclusion fails: $f(x)=\tfrac{1}{x^{2}}$ on $[1,\infty)$ is monotone, nonnegative and bounded, and its integral converges to $1$. Drop monotone and it also fails: a nonnegative unbounded function made of ever-taller, ever-thinner spikes (height $n$ and width $2^{-n}n^{-1}$ near $x=n$) has total area $\sum 2^{-n}<\infty$. So both conditions genuinely do work here.

6. 6

   Note the contrast with series. The analogous statement for sequences is trivial — an unbounded nonnegative sequence cannot have $a_n\to 0$, so $\sum a_n$ diverges by the term test. For integrals no such term test exists (a function can fail to tend to $0$ and still have a convergent integral), which is why monotonicity has to be assumed explicitly.