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Thread: Definite integral

  1. #1
    Member roshanhero's Avatar
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    Definite integral

    Integrate:
    $\displaystyle \int_{-\infty }^{\infty}\frac{x^2}{1+x^4}$
    Please give me the hints only
    Last edited by roshanhero; Jun 9th 2010 at 09:04 PM. Reason: My mistake
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  2. #2
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    ....

    First show

    $\displaystyle \int_0^{\infty} \frac{x^2}{1 + x^4}~dx $ converges

    Then we can write

    $\displaystyle \int_{-\infty}^{\infty} \frac{x^2}{1 + x^4}~dx = 2 \int_0^{\infty}\frac{x^2}{1+x^4}~dx $

    For the integral $\displaystyle 2 \int_0^{\infty}\frac{x^2}{1+x^4}~dx $

    Write the numerator as $\displaystyle \frac{1}{2} ( (x^2+1)+(x^2-1)) $

    Then it equals to :

    $\displaystyle \int_0^{\infty}\frac{(x^2+1)+(x^2-1)}{1+x^4}~dx $

    $\displaystyle =\int_0^{\infty}\frac{x^2+1}{1+x^4}~dx + \int_0^{\infty}\frac{x^2-1}{1+x^4}~dx $


    For the integral $\displaystyle \int_0^{\infty} \frac{x^2+1}{1+x^4}~dx $

    $\displaystyle = \int_0^{\infty} \frac{1+1/x^2}{1/x^2+x^2}~dx $

    $\displaystyle = \int_0^{\infty} \frac{d(x-1/x)}{ \left( x - \frac{1}{x} \right)^2 + 2 }$ then use substitution : $\displaystyle x-1/x = t $ ...

    But the second integral should be zero : $\displaystyle I = \int_0^{\infty}\frac{x^2-1}{1+x^4}~dx$

    After substituting $\displaystyle x = 1/t $ it becomes

    $\displaystyle I = \int_0^{\infty} \frac{1- t^2}{1+t^4}~dt = - \int_0^{\infty} \frac{ t^2-1}{1+t^4}~dt = -I ~~~\implies I = 0$
    Last edited by simplependulum; Jun 9th 2010 at 09:20 PM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The complex analysis offers a faster way to arrive to the result. If we apply the residue theorem we have...

    $\displaystyle \int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{4}}\cdot dx = 2 \pi i \sum_{k} r_{k}$ (1)

    ... where the $\displaystyle r_{k}$ are the residue of the poles of $\displaystyle f(z) = \frac{z^{2}}{1+z^{4}}$ with positive imaginary part. If $\displaystyle z_{0}$ is a pole of $\displaystyle f(z)$ is [applying l'Hopital rule] ...

    $\displaystyle r_{z=z_{0}} = \lim_{z \rightarrow z_{0}} (z-z_{0}) f(z) = \frac{1}{4 z_{0}}$ (2)

    Now the poles of $\displaystyle f(z)$ with positive imaginary part are $\displaystyle z= \frac{i \pm 1}{\sqrt{2}}$ so that is...

    $\displaystyle \int_{-\infty}^{+\infty} \frac{x^{2}}{1+x^{4}}\cdot dx = 2 \pi i \frac{\sqrt{2}}{4} (\frac{1}{i-1} + \frac{1}{i+1}) = \frac{\pi}{\sqrt{2}} $ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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