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

  1. Aug 1, 2014 #1
    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 dont 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.

    Attached Files:

    Last edited: Aug 1, 2014
  2. jcsd
  3. Aug 2, 2014 #2


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    Homework Helper

    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.
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