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I can't use any of the convergence tests I learned because I have no information on $f(x)$, in particular I don't know if it's continuous or positive.

The only thing I could think of was that if $\displaystyle \int_{1}^{\infty}f(x)\ \mathrm dx$ was absolutely convergent, then $|f(x)\sin x| \leq |f(x)|$ would imply by the comparison test that $\displaystyle \int_{1}^{\infty}f(x)\sin x\ \mathrm dx$ converges.

So if I want to find a counter-example I have to pick $f(x)$ so that $\displaystyle \int_{1}^{\infty}f(x)\ \mathrm dx$ conditionally converges, but I can't think of one.

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Consider $f(x)=\sin(x) / x$.

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    I showed $\int_1^\infty {\frac{{\sin (x)}}{x}} {\rm d}x$ converges using Dirichlets test and divergence using $\sin^2 (x) = (1-\cos(2x))/2$ like you said.2010-12-02