Ngô Quốc Anh

February 9, 2010

CE: Integral calculus

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


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


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.



  1. It seems that the improper integral

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

    should be understood as Riemann improper integral, since your counter example

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

    divergence in sense of Lebesgue improper integral.

    Comment by mrstoke — April 29, 2012 @ 22:22

    • Yes, all integrals should be understood as Riemann improper integrals. It would be interesting if you could provide a proof of the following fact: the integral \int_0^\infty \frac{\sin x}{x}dx diverges in the sense of Lebesgue improper integrals.

      Comment by Ngô Quốc Anh — April 29, 2012 @ 22:29

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