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Explore absolute and conditional convergence of the integral $$\int_{1}^{+\infty} \frac{\sin \sqrt[3] x}{x-\ln x}\, dx$$ My general ideas are the following:

  • for absolute convergence I should use comparison test and therefore find second function but my attempts didn't work.
  • for conditional convergence I should use Dirichlet's test where I take $f(x) = \sin \sqrt[3] x$ and $g(x) = \frac{1}{x-\ln x}$. Are all conditions to use the test met?
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  • $\begingroup$ $f(x)$ is not a decreasing sequence, so I don't think Dirichlet's test will work. $\endgroup$ Commented Apr 11, 2019 at 21:33
  • $\begingroup$ You have $1/(x - \ln x) \to 0$ monotonically, but $\int_1^x \sin t^{1/3} \, dt$ is not bounded for all $x > 1$ so the Dirichlet test will not work directly. $\endgroup$ Commented Apr 12, 2019 at 0:19

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For conditional convergence note that

$$\int_1^c \frac{\sin x^{1/3}}{x- \ln x}dx = \int_1^{c^{1/3}} \frac{\sin u}{u^3- 3\ln u}3u^2 \, du = 3\int_1^{c^{1/3}} \frac{\sin u}{u- 3u^{-2}\ln u} \, du$$

The integral on the RHS is convergent by Dirichlet's test since $(u- 3u^{-2}\ln u)^{-1} \to 0$ monotonically as $u \to \infty$.

To prove that the integral is not absolutely convergent exploit the periodicity of $\sin$ by decomposing into a divergent sum of integrals over intervals of the form $[k\pi, (k+1)\pi]$. Proceed as is done for the well-known proof that $\int_1^\infty x^{-1} \sin x \, dx$ is not absolutely convergent.

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