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Convergence of improper integrals

  1. Jun 8, 2015 #1
    What is the difference between

    [itex]\int_{-\infty}^{\infty} \frac{x}{1+x^2}dx[/itex]

    and

    [itex]\lim_{R\rightarrow \infty}\int_{-R}^{R} \frac{x}{1+x^2}dx[/itex] ?

    And why does the first expression diverge, whilst the second converges and is equal to zero?
     
  2. jcsd
  3. Jun 8, 2015 #2

    PeroK

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    Welcome to Physics Forums!

    The first is an improper integral and is defined as:

    [itex]\int_{-\infty}^{\infty} \frac{x}{1+x^2}dx = \int_{-\infty}^{0} \frac{x}{1+x^2}dx + \int_{0}^{\infty} \frac{x}{1+x^2}dx [/itex]

    In other words, for the improper integral to be finite both positive and negative parts must be finite.

    The second is called the "Cauchy Principal Value". You can google for this. Odd functions like yours or like ##sin(x)## will have CPV = 0, even though the improper integral is not finite.
     
  4. Jun 8, 2015 #3
    I apologise in advance if this is a stupid question, but could one say that the two expressions are different because the infinities in the first expression may or may not be of the same magnitude, and so the area under curve is therefore undefined. Whereas the infinities in the second expression must be of the same magnitude, and so the areas either side of the y-axis must therefore be equal, leading them to cancel out as in an odd function?
     
  5. Jun 8, 2015 #4

    PeroK

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    They don't need to be infinities. Take the example of ##sin(x)## or ##cos(x)##: the improper integral is not well defined because the integrals are oscillating functions and never settle down to a finite value, although the integrals themselves are bounded.
     
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