# Integral $\int _1^{\infty }\sin^2 \left(\frac{3}{x} \right)dx$

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?

• You can only separate integrands when both converge. e.g. $\int 0 dx = \int 1 dx - \int 1 dx$ doesn't hold – 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 $$\alpha>1.$$
• 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 – Sagigever May 13 '20 at 10:50
• @Sagigever $\sin x \leq x$ for all $x\geq 0.$ – Sahiba Arora May 13 '20 at 10:51