I need to determine if the following integral diverges or converges, and if it is convergent then whether it is absolutely convergent. $$\int _1^{\infty }\sin^2 \left(\frac{3}{x} \right)dx$$

I used the formula of $\cos(2\alpha)=1-2\sin^2(\alpha)$ hence I get to : $$\int _1^{\infty }\sin^2 \left(\frac{3}{x}\right)dx=\int _1^{\infty }\frac{1}{2}dx - \frac{1}{2}\int _1^{\infty }\cos\left(\frac{6}{x}\right)$$

but $\int _1^{\infty }\frac{1}{2}dx$ is divergent so how is it possible that $\int _1^{\infty }\sin^2\left(\frac{3}{x}\right)dx$ converges if it has one divergent integral in its sum? Am I doing something wrong in my assumption?

  • 6
    $\begingroup$ You can only separate integrands when both converge. e.g. $\int 0 dx = \int 1 dx - \int 1 dx$ doesn't hold $\endgroup$ – David P May 13 '20 at 10:27

As pointed out by David Peterson in the comments, the error you make is splitting $$\frac{1}{2}\int_1^{\infty}1-\cos\left(\frac{6}{x}\right)\ dx= \int _1^{\infty }\frac{1}{2}dx - \frac{1}{2}\int _1^{\infty }\cos\left(\frac{6}{x}\right)\ dx$$ which you can only do when both integrals on the RHS converge.

In fact, the original improper integral is convergent. Since $$0\leq \sin^2\left(\frac{3}{x}\right)\leq \frac{9}{x^2}$$ and $$\int_1^{\infty}\frac{9}{x^2}\ dx$$ converges so does $$\int_1^{\infty} \sin^2\left(\frac{3}{x}\right) \ dx$$ by the comparison test. In fact $$\int_1^{\infty} \sin^{\alpha}\left(\frac{1}{x}\right)\ dx$$ converges for every $\alpha>1.$

  • $\begingroup$ Can you explain plz the $0\leq \sin^2\left(\frac{3}{x}\right)\leq \frac{9}{x^2}$, not sure I understood the right side of the expression $\endgroup$ – Sagigever May 13 '20 at 10:50
  • 3
    $\begingroup$ @Sagigever $\sin x \leq x$ for all $x\geq 0.$ $\endgroup$ – Sahiba Arora May 13 '20 at 10:51

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.