How to show $\int_{1}^{\infty} \frac{\sin^2(x)}{x^2}dx$ is finite?

Question

At first, my approach was to directly take the improper integral of it.

However, it seems not that easy.

Then I tried to find another fraction to make a comparison. I got $\frac{\sin^2(x)}{x^2} < \frac{\sin^2(x)}{x}$. So if I could show $\int_{1}^{\infty} \frac{\sin^2(x)}{x}dx $ is finite，$\int_{1}^{\infty} \frac{\sin^2(x)}{x^2}dx$ will then be finite . Nevertheless, I still cannot figure the latter out.

Could anyone suggest me what function to compare to, or other method?

Thank you!

Answer 1

Hint: compare with the integral $1/x^2$

Answer 2

It helps to notice that the $\sin(x)$ part shouldn't impact the final answer because your indefinite integral goes to infinite and $\sin(x)$ will most likely cancel itself out in the process. Simply put, as Tsemo Aristide noted, the indefinite integral for $\frac1{x^2}$ should do.

You could also take the following method:

$$\int_1^{\infty}\frac{\sin^2(x)}{x^2}dx=\int_1^{\infty}\sin^2(x)\cdot\frac1{x^2}dx$$

Perform integration by parts infinitely, then use the fundamental theorem of Calculus. Since we have a $\frac1{x^n}$, then the entire thing will go to $0$ as $x$ goes to infinity. Also, $\frac1{1^n}=1$, taking care of the second part in the indefinite integral part.

$$\int\sin^2(x)\cdot\frac1{x^2}dx=-\sin^2(x)x^{-1}+2\int\cos(x)\sin(x)(x^{-1})dx$$

At this point, you would go and look up integration by parts with 3 terms.

Answer 3

Since $0\leqslant \sin^2x\leqslant 1$ for all $x\in \mathbb{R}$, then we can use following inequality $$0<\int_1^\infty \frac{\sin^2x}{x^2}dx\leqslant \int_1^\infty \frac{dx}{x^2}=1.$$

Answer 4

This is not an answer but it is too long for a comment.

You received answers showing that bounding the integral is quite simple.

If you already know special functions (as Brevan Ellefsen commented) , you can even do more.$$I=\int\frac{\sin^2(x)}{x^2}dx=\frac 12\int\frac{1-\cos(2x)}{x^2}dx=\frac 12\int\frac{dx}{x^2}-\frac 12\int\frac{\cos(2x)}{x^2}dx$$ The first integral does not present any problem.

Now, consider $$J=\int\frac{\cos(2x)}{x^2}dx$$ and integrate by parts $$u=\cos(2x) \quad , \quad du=-2\sin(2x)dx \quad, \quad dv=\frac{dx}{x^2}\quad, \quad v=-\frac{1}{x}$$ $$J=-\frac{\cos (2 x)}{x}-2\int \frac{ \sin (2 x)}{x}dx=-\frac{\cos (2 x)}{x}-2\int \frac{ \sin (2 x)}{2x}d(2x)$$ So, $$J=-\frac{\cos (2 x)}{x}-2\,\text{Si}(2 x)$$ where appears the definition of the sine integral function. Finally, $$I=\text{Si}(2 x)+\frac{\cos(2x)-1}{2x}$$

Now, using the given bounds, $$\int_{1}^{\infty} \frac{\sin^2(x)}{x^2}dx=\frac{1}{2} (1+\pi -\cos (2))-\text{Si}(2)\approx 0.673457$$