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Evaluating integral from -∞ to ∞

  1. Mar 17, 2013 #1
    When evaluating an integral [tex]\int^∞_{-∞}[/tex] do I have to split the integral up like [tex]\int^∞_0 + \int^0_{-∞}[/tex] and evaluate it this way or can I simply keep the integral in it's original form and plug in infinity and negative infinity into the expression and evaluate it?
  2. jcsd
  3. Mar 17, 2013 #2


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    It depends on the expression you get for the indefinite integral. In practice most of the integrals of this nature require more than getting an antiderivative.
  4. Mar 17, 2013 #3
    Well right now I am just learning how to solve improper integrals. I was working on [tex]\int^∞_{-∞} \frac{1}{2}tan^{-1} x^2 [/tex]. I don't know if I am doing this right but splitting up and keeping the integral together gets me two different answers. Splitting it up gets me the right answer
  5. Mar 17, 2013 #4
    Since it's an even function, the property [tex] \displaystyle \int_{-a}^a{ f(x)\,dx } = 2\int_0^a{f(x)\,dx}[/tex] holds.

    So that means you can evaluate it as [tex]\displaystyle 2\int_0^{\infty}{\frac{1}{2}\arctan{\left( x^2 \right)} \,dx} [/tex].
  6. Mar 18, 2013 #5


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    Please tell us what two different answer you get.

    You should know that [itex]\int_0^\infty f(x)dx[/itex] is defined as [itex]\lim_{a\to\infty}\int_0^a f(x) dx[/itex] while [itex]\int_{-\infty}^\infty f(x)dx[/itex] is defined as [itex]\lim_{a\to -\infty}\lim_{b\to\infty} \int_a^b f(x)dx[/itex].

    In particular, you cannot use the same variable, and the same limits for the two limits of integration. That is, the "Cauchy Principle Value",
    [tex]\lim_{a\to\infty} \int_{-a}^a f(x)dx[/tex]
    may exist when the integral itself does not.
  7. Mar 19, 2013 #6
    OK that's why I have been getting the wrong answer. I was using one variable as both limits. Thanks for the answer.
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