I have the following two integrals:

$$I_1 = \int_0^\infty \frac{\sin^2(x)}{x^2}dx$$

$$I_2 = \int_0^\infty \frac{\cos^2(x)}{x^2}dx$$

I don't understand why $I_1$ converges and $I_2$ diverges. I tried $\frac{\sin^2(x)}{x^2} \le \frac{1}{x^2}$ but that does not work since $\int \frac{1}{x^2} dx$ diverges. I am convinced that one of them has to diverge since $I_1+I_2$ diverges, but I don't see why.

  • 1
    $\begingroup$ Try graphing each of the two functions near $x=0$. For, say, $0 \le x \le 1$, you not only have $\frac{\sin^2 x}{x^2} \le \frac{1}{x^2}$ but even $\frac{\sin^2 x}{x^2} \le 1$. $\endgroup$ Oct 27, 2016 at 17:52

1 Answer 1


There are two problems : one in $0$ and one in $+\infty$. Both functions are integrable on an interval of the form $\left[a,+\infty\right[$ with $a>0$.

But around $0$, $\frac{\sin^2x}{x^2}\sim 1$ is integrable on $\left[0,a\right]$, but $\frac{\cos^2x}{x^2}\sim \frac{1}{x^2}$ which is not.


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