# if $\int_1^{\infty}f(x)\ \mathrm dx$ converges, must $\int_1^{\infty}f(x)\sin x\ \mathrm dx$ converge?

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.

Consider $f(x)=\sin(x) / x$.
• 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. – Shai Covo Dec 2 '10 at 14:43
• 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. – Willie Wong Dec 2 '10 at 16:31
• I didn't want to give too many hints, but showing divergence is very easy if you use $\sin^2 (x) = (1-\cos(2x))/2$. – Shai Covo Dec 2 '10 at 17:45
• 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. – daniel.jackson Dec 2 '10 at 20:28