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?
  • $\begingroup$ $f(x)$ is not a decreasing sequence, so I don't think Dirichlet's test will work. $\endgroup$ – D.B. Apr 11 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. $\endgroup$ – RRL Apr 12 at 0:19

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.