Improper integral comparison test

[hitemup]

The question asks whether the following converges or diverges.

\int_{0}^{\infty } \frac{\left | sinx \right |}{x^2} dx

Now I think there might be a trick with the domain of sine function but I couldn't make up my mind on this.
I tried to compare it with 1/x^2, (sinx)/x, and sinx. I actually expected that I would get something good with 1/x^2, but as the lower limit of the integral is zero, it ended up with infinity on (0, inf) and since 1/x^2 is greater than (sinx)/x^2, and is divergent as we just found, we cannot say whether the given function diverges or converges. So I'm wondering what is the right track on this question?

 

[Stephen Tashi]

\int_{0}^{\infty } \frac{\left | sinx \right |}{x^2} dx =

\int_{0}^{ \frac{\pi}{2} } \frac{\left | sinx \right |}{x^2} dx + \int_{ \frac{\pi}{2} }^{\infty} \frac{\left | sinx \right |}{x^2} dx

Deal with the two integrals separately.

In the interval [0,\frac{\pi}{2}] you can establish an inequality between |\sin(x)| and a non-periodic function. Try comparing it to f(x) = x.