1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook