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Integrals to infinity

  1. Jan 22, 2010 #1
    Hi all

    If I have an integral from -∞ to ∞, then is it always true that we can write it as a limit? I.e. if we have a continuous function f, then is it always true that

    [tex]
    \int_{ - \infty }^\infty {f(x)dx = \mathop {\lim }\limits_{N \to \infty } \int_{ - N}^N {f(x)dx} }
    [/tex]
    ?
     
  2. jcsd
  3. Jan 22, 2010 #2
    No, the lower bound is independent of the upper bound, it is actually:

    [tex]lim_{N -> \infty, M -> -\infty}\int^{N}_{M}f(x)dx[/tex]

    But if this limit exists then it equals the expression as you wrote it.

    Your limit is called the principal value (p.v.) of the integral.
     
  4. Jan 22, 2010 #3
    Thanks for the swift response. What if I have e.g.

    [tex]
    \int_{ - \infty}^\infty {f(x)\sin xdx}
    [/tex]
    where I know that f is an even function? I wish to argue that the integral is zero since it runs over a symmetric interval, but I am not sure if - ∞..∞ is a symmetric interval? That is why I tried writing the limits, but if they are independent, then I am not quite sure my argument holds.
     
    Last edited: Jan 22, 2010
  5. Jan 22, 2010 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    The argument holds as long as |f(x)| is integrable over the entire real line. If not, then the integral you want is indeterminate.
     
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