Ngô Quốc Anh

September 4, 2010

CE: Convergence of improper integrals does not imply the integrands go to zero

Filed under: Counter-examples, Giải Tích 3 — Ngô Quốc Anh @ 12:31

This entry devotes a similar question that raises during a course of series. We all know that for a convergent series of (positive) real number

\displaystyle \sum_{n=1}^\infty a_n

it is necessary to have

\displaystyle\mathop {\lim }\limits_{n \to \infty } {a_n} = 0.

This is the so-called n-th term test. A natural extension is the following question

Question. Suppose f(x) is positive on [0,\infty) and

\displaystyle\int_0^{ + \infty } {f(x)dx}

exists. Must f(x) tend to zero as x \to +\infty?

I will show that this is indeed not the case. Take

\displaystyle f(x)=\frac{1}{e^x}+g(x), \quad x \geqslant 0


\displaystyle g(x)=\begin{cases}0, & x \in \left[0,\frac{7}{4}\right], \\ 1, & x=2,3,..., \\ n^2(x-n)+1,& x \in [n-n^{-2},n], n=2,3,..., \\ -n^2(x-n)+1,& x\in [n,n+n^{-2}],n=2,3,..., \\ 0,& \text{otherwise}.\end{cases}

The above solution tell us the fact that the convergence of improper integral \int_0^{ + \infty } {f(x)dx}  is not enough to guarantee \mathop {\lim }_{x \to \infty } f(x) = 0.

It turns out to propose the question under what condition on f, f(x) tend to zero as x \to +\infty? There are several answers in the literature, for example

  • If f is positive, differentiable and |f'(x)| is uniformly bounded, say by M, then by the mean value theorem, we can prove \mathop {\lim }_{x \to \infty } f(x) = 0.
  • If f is uniformly continuous on [0,\infty) then the same conclusion still holds by the Cauchy theorem.
  • If f is monotone decreasing, we obtain a stronger result, i.e. \mathop {\lim }_{x \to \infty } xf(x) = 0.

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