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

[Mark_M]

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

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


The magnitude became smaller and smaller, since [tex]\sin x[/tex] 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
[tex]\int_{2.04}^\infty \frac{\sin x}{x^2}dx=\frac{\sin(2.04)}{2.04}-Ci(2.04)[/tex], where [tex]Ci(x)[/tex] is the cosine integral function.

 

[pasmith]

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

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


The magnitude became smaller and smaller, since [tex]\sin x[/tex] 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
[tex]\int_{2.04}^\infty \frac{\sin x}{x^2}dx=\frac{\sin(2.04)}{2.04}-Ci(2.04)[/tex], where [tex]Ci(x)[/tex] is the cosine integral function.

So all you need to know is whether or not [tex]\int_{2.04}^{2\pi} \frac{\sin x}{x^2}\,dx \geq 0.[/tex] 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.