Could you, please, give any reasons why this integral converges. I have tried a lot of different methods (e.g. Dirichlet's test) none of them helped me

$$\int\limits_1^\infty \frac{x\sin x}{x^2+3x+3}\,dx$$

  • $\begingroup$ Is contour integration (complex) allowed? $\endgroup$ – imranfat May 28 '18 at 17:38
  • $\begingroup$ @imranfat No, it isn’t $\endgroup$ – FoRRestDp May 28 '18 at 17:39
  • $\begingroup$ Well, then the provided answer is also a little issue (it is not incorrect), because that assumes convergence of the integral involving $sinx/x$, so that needs to be addressed first $\endgroup$ – imranfat May 28 '18 at 17:40
  • $\begingroup$ Wolfram gives a numerical result of it here $\endgroup$ – TheSimpliFire May 28 '18 at 17:49
  • 1
    $\begingroup$ What was the trouble applying Dirichlet test? $\endgroup$ – A.Γ. May 28 '18 at 18:24

Hint: if you know that the integral of $\frac{\sin x}{x}$ is convergent, you can write $$ \frac{\sin x}{x}=\frac{x^2+3x+3}{x^2+3x+3}\frac{\sin x}{x}=\frac{x\sin x}{x^2+3x+3}+\frac{3\sin x}{x^2+3x+3}+\frac{3\sin x}{(x^2+3x+3)x} $$ and combine the terms to express your integral of interest via integrals that are easier to prove convergence for.

P.S. If you don't then try integration by parts (integrate $\sin x$ and derivate the rational part).


You can apply a Dirichlet test for convergence of improper integrals


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.