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Improper integral from -infinity to infinity

  1. Mar 2, 2012 #1
    What is the improper integral of sin[x]/(1+x^2) from -infinity to infinity equal to? It doesn't seem to be integrable in the real number system.

    Is it 0, because sin[x] is an odd function and 1+x^2 is an even function, and an odd function divided by an even function is equal to an odd function which has a rule that f(x) + f(-x) = 0?
     
  2. jcsd
  3. Mar 3, 2012 #2

    jbunniii

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    Why do you say that? We have

    [tex]\left|\frac{\sin(x)}{1 + x^2}\right| \leq \frac{1}{1 + x^2}[/tex]

    and the function on the right has a finite integral on the interval [itex](-\infty,\infty)[/itex]. (Indeed, the integral equals [itex]\pi[/itex].) Therefore

    [tex]\left|\int_{-\infty}^{\infty}\frac{\sin(x)}{1 + x^2} dx \right| \leq \int_{-\infty}^{\infty} \left|\frac{\sin(x)}{1 + x^2}\right| dx \leq \pi[/tex]

    Not all odd functions have zero integral. E.g. [itex]\int_{-\infty}^{\infty}\sin(x)dx[/itex] doesn't converge. However, any odd function for which [itex]\int_{0}^{\infty}f(x)dx[/itex] is finite will satisfy

    [tex]\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{0} f(x) dx + \int_{0}^{\infty}f(x)dx = \int_{0}^{\infty}f(x)dx - \int_{0}^{\infty}f(x)dx = 0[/tex]

    So see if you can show that [itex]\int_{0}^{\infty}\sin(x)/(1 + x^2) dx[/itex] is finite, using a similar argument to what I wrote above.
     
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