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Math Help - improper integral

  1. #1
    Member disclaimer's Avatar
    Joined
    Dec 2007
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    77

    improper integral

    Hi all,

    Maybe a stupid question, but how to calculate the following integral:
    \int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}
    I would like to see what is the method of calculating such integrals, because I don't quite get what is written on Wikipedia:

    According to the formula above, I assume it's:
    \int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\lim_{t\rig  htarrow-\infty}\int_{t}^{a}{\frac{dx}{x^2+2x+2}}+\lim_{t\r  ightarrow+\infty}\int_{a}^{t}{\frac{dx}{x^2+2x+2}}
    Then I want to use substitution x+1=u, but I am not sure how the integration limits will change; is it a+1 while t remains intact? If so, then it would be:
    \lim_{t\rightarrow-\infty}\int_{t}^{a+1}{\frac{du}{u^2+1}}+\lim_{t\ri  ghtarrow+\infty}\int_{a+1}^{t}{\frac{du}{u^2+1}}
    Does it further give something like \lim_{t\rightarrow-\infty}\left(\arctan{(a+1)}-\arctan{t}\right)+\lim_{t\rightarrow+\infty}\left(  \arctan{t}-\arctan{(a+1)}\right)?
    And, finally, \int_{-\infty}^{+\infty}{\frac{dx}{x^2+2x+2}}=\pi
    I have no idea, that was just my guess.... Thanks for any input.
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  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
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    It's ok , you didn't need to change the t because it's going to infinity.

    But: \int_a^b\frac{dx}{x^2+2x+2}=\int_{a+1}^{b+1}\frac{  du}{u^2+1}
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