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Given that $f:[0, \infty] \to \mathbb{R}$ is decreasing with $\displaystyle\lim_{x \rightarrow \infty} f(x)=0$, prove that

$$I=\int_{0}^{1}\frac{\cos(\frac{1}{x})f(\frac{1}{x})}{x^2}dx$$ converges.

I've thought about using Dirichlet test but it works only if $f$ is continuously differentiable. It can be an improper integral or not, depends how $f$ is defined, so some of my other ideas didn't work either. Any ideas?

Rodrigo de Azevedo
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35T41
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  • You can use Dirichlet's test as it is formulated here: https://en.wikipedia.org/wiki/Dirichlet%27s_test#Improper_integrals, which doesn't require $f$ to be uniformly continuous. – Micapps Jun 26 '16 at 11:46
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    Substitute $u=1/x$ and the integral becomes $\int_1^{\infty} \cos(u)f(u)du$ and the theorem cited in this solution should cover it: http://math.stackexchange.com/questions/141048/dirichlets-test-for-convergence-of-improper-integrals – Lance Sackless Jun 26 '16 at 11:46
  • @LanceSackless Are you sure that $f \in C^1$? – Marco Lecci Jun 26 '16 at 12:21

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By Lebesgue differentiation theorem, a decreasing function is almost everywhere differentiable, so for any $\varepsilon >0$ $$ \int_{\varepsilon}^{1}\frac{f(1/x)\cos(1/x)}{x^2}\,dx = \int_{1}^{1/\varepsilon}\cos(x)\,f(x)\,dx$$ and we may apply integration by parts/Dirichlet's test. It does not really matter that $f'(x)$ is not defined at some points, and we do not need the continuity of $f'$.

Jack D'Aurizio
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