Integral -- analytical way to prove this integral is non-negative?

[Mark_M]

Is there any analytical way to prove that the integral \int_{2.04}^\infty \frac{\sin x}{x^2}dx is nonegative?

I tryed to use geometrical approach, i.e. the graph of the integrand look like:


The magnitude became smaller and smaller, since \sin x is multiplied by the decreasing function $1/x^2$, so the third area, which is positive, is bigger then the forth one, which is negative, and so on. BUT I don't know what to do with the first two areas(


OR integration by parts gave me
\int_{2.04}^\infty \frac{\sin x}{x^2}dx=\frac{\sin(2.04)}{2.04}-Ci(2.04), where Ci(x) is the cosine integral function.

 

[pasmith]

Is there any analytical way to prove that the integral \int_{2.04}^\infty \frac{\sin x}{x^2}dx is nonegative?

I tryed to use geometrical approach, i.e. the graph of the integrand look like:


The magnitude became smaller and smaller, since \sin x is multiplied by the decreasing function $1/x^2$, so the third area, which is positive, is bigger then the forth one, which is negative, and so on. BUT I don't know what to do with the first two areas(


OR integration by parts gave me
\int_{2.04}^\infty \frac{\sin x}{x^2}dx=\frac{\sin(2.04)}{2.04}-Ci(2.04), where Ci(x) is the cosine integral function.

So all you need to know is whether or not \int_{2.04}^{2\pi} \frac{\sin x}{x^2}\,dx \geq 0. Evaluate it numerically.

If that doesn't satisfy your "analytic way" criterion, then you can instead show that there exists a lower Darboux sum which is strictly positive. It then follows that the integral is strictly positive.