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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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    $\begingroup$ To show $\int_1^\infty {\frac{{\sin (x)}}{x}} {\rm d}x$ exists, split the integral over intervals of the form $[k\pi,(k+1)\pi]$, $k \in \mathbb{N}$, and consider the alternating series test. To show $\int_1^\infty {\frac{{\sin^2 (x)}}{x}} {\rm d}x$ diverges, split the integral as before, and consider comparison with the harmonic series. $\endgroup$ – Shai Covo Dec 2 '10 at 14:43
  • $\begingroup$ As an aside, one way to approximate the value of an oscillatory integral is to treat it as an alternating series, where the terms of the infinite sum are the integrals between consecutive zeroes or extrema, and then use numerical methods for summing alternating series. $\endgroup$ – J. M. is a poor mathematician Dec 2 '10 at 16:21
  • $\begingroup$ I think a modification to your example for which it is easier to show convergence/divergence is to take $f(x) = -1^n / n$ when $\pi (n-1) \leq x < \pi n$. Then $\int f dx$ and $\int f\sin dx$ are, up to a small constant correction from the first segment between $(1,\pi)$, precisely the alternating and non-alternating harmonic series. $\endgroup$ – Willie Wong Dec 2 '10 at 16:31
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    $\begingroup$ I didn't want to give too many hints, but showing divergence is very easy if you use $\sin^2 (x) = (1-\cos(2x))/2$. $\endgroup$ – Shai Covo Dec 2 '10 at 17:45
  • $\begingroup$ 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. $\endgroup$ – daniel.jackson Dec 2 '10 at 20:28

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