Ngô Quốc Anh

September 4, 2010

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?

(more…)

February 9, 2010

CE: Integral calculus

Filed under: Các Bài Tập Nhỏ, Counter-examples, Giải Tích 1, Giải Tích Cổ Điển — Ngô Quốc Anh @ 22:24

Question: If the integral

\displaystyle \int_0^\infty f (x)dx

converges and a function y = g(x) is bounded then the integral

\displaystyle \int_0^\infty f (x)g(x)dx

converges.

Observation: It seems since g(x) is bounded by a constant called M, the integral \int_0^\infty f (x)g(x)dx is then dominated by M times \int_0^\infty f (x)dx which becomes a finite number.

Counter-example: The integral

\displaystyle \int_0^\infty \frac{\sin x}{x}dx

converges and the function g(x)=\sin x is bounded but the integral

\displaystyle \int_0^\infty \frac{\sin^2 x}{x}dx

diverges.

Explanation: What we thought is the following estimate

\displaystyle\left| {\int_0^\infty {f(x)g(x)dx} } \right| \leqslant \int_0^\infty {\left| {f(x)g(x)} \right|dx} \leqslant M\int_0^\infty {|f(x)|dx}.

However, the convergence of \int_0^\infty f (x)dx is not sufficient to imply that \int_0^\infty |f (x)|dx<\infty.

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