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Evaluation of improper integral involving sinx/x

  1. Feb 2, 2013 #1
    Sir,

    Recently when i am evaluating a convolution integral, i came across the integral of |sinx/x| under limits running from 0 to infinity. when i tried to evaluate the integral, i used complex analysis tools like assuming a function e^(iz) / z and deduce the above integral from integral
    of f(z) by contour integration.But, i am unable to arrive at any conclusion.I wanted to know whether the integral is finite or not.Can you please help me.
     
  2. jcsd
  3. Feb 2, 2013 #2
    Do you mean the integral with or without the absolute value sign in the integrand? You use the absolute value in one place but not the other.

    Without the absolute value, this is a Dirichlet integral. Evaluating it with a few different methods (including contour integration) is shown in the link.
     
  4. Feb 2, 2013 #3

    jbunniii

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    With the absolute value, the integral is infinite. To see this, observe that
    $$\left|\frac{\sin(x)}{x}\right| > \left|\frac{1}{2x}\right|$$
    if and only if
    $$\left|\sin(x)\right| > \frac{1}{2}$$
    if and only if
    $$x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] + n\pi$$
    where ##n## is any integer. Therefore,
    $$\int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right| dx \geq \sum_{n=0}^{\infty}\int_{\frac{\pi}{6} + n\pi}^{\frac{5\pi}{6}+n\pi} \left|\frac{1}{2x}\right| dx \geq \sum_{n=0}^{\infty} \frac{2\pi}{3} \left(\frac{1}{\frac{\pi}{3} + 2n\pi}\right)$$
    and the rightmost expression is infinite by comparison with a harmonic series.
     
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