# 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

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

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 $\int_0^\infty |f (x)|dx<\infty$.

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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