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Math Help - difficult Integration

  1. #1
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    difficult Integration

    Calculate

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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by dapore View Post
    Calculate

    The only thing I can say is that your lovely integral is "absolute convergent" ...


    It's hard to believe that someone sane will ask you to compute this...
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  3. #3
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    It is "good" integral
    at x=0 =0
    at x=inf decreases as 1/x^2
    and its value is
    integrate sin x dx/ (1+x^2) from x=0 to +inf - Wolfram|Alpha
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  4. #4
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    Thank you I thought in the following way :
    By integration by parts, we know that 2\int_{0}^{\infty}a e^{-a^{2}(1+x^{2})}da =\frac{1}{1+x^{2}}

    We then get:
     I =\int_{0}^{\infty}\frac{\sin x}{1+x^{2}}dx = 2\int_{0}^{\infty}ae^{-a^{2}}\boxed{\int_{0}^{\infty}e^{-a^{2}x^{2}}\sin x dx}da
    by working out the boxed part using integration by parts we get the answer


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  5. #5
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    Quote Originally Posted by dapore View Post
    by working out the boxed part using integration by parts we get the answer.
    And how do you propose that we work out the boxed part? I would be interested to know.
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  6. #6
    Newbie Bourbaki's Avatar
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    Quote Originally Posted by dapore View Post
    Calculate

    Use the Residue Theorem.
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  7. #7
    MHF Contributor chisigma's Avatar
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    Who hasn't at disposal Wolfram's tools but only a table of Laplace Tranforms can remember that is...

    \displaystyle \varphi(s)= \mathcal{L} \{\frac{1}{1+t^{2}}\} = \int_{0}^{\infty} \frac{e^{-s\ t}}{1+t^{2}}\ dt = \cos  s \ \{\frac{\pi}{2} - Si (s) \} - \sin s \ Ci(s) (1)

    ... so that is...

    \displaystyle \int_{0}^{\infty} \frac{\sin t}{1+t^{2}} \ dt = Im \{ \varphi(-i)\} (2)

    Kind regards

    \chi \sigma
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Wow! Chinese!
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