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Math Help - Residue theorem

  1. #1
    Junior Member
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    Dec 2009
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    Residue theorem

    Hi I'm having some problems with the below question

    Use the residue theorem to compute

    \int\limits_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx


    So i rewrote the intergral as

    \int\limits_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx = \Re(\int\limits_{-\infty}^{\infty} \frac{e^{jx}}{1+x^2} dx)

    Where do i go from here?

    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    The value of the complex integral...

    \int_{-\infty}^{+\infty} \frac{e^{j x}}{1+x^{2}}\cdot dx (1)

    ... is given by...

    \lim_{R \rightarrow \infty} \int_{\Gamma} f(z)\cdot dz (2)

    ... where f(z) = \frac{e^{j z}}{1+z^{2}} and \Gamma is the red path represented in figure...




    All that is true only if the socalled 'Jordan's lemma' is verified, just as in this case. For the 'second Cauchy integral formula' is...

    \int_{\Gamma} f(z)\cdot dz = 2 \pi j \sum_{i} r_{i} (3)

    ... where r_{i} is the residue relative at each pole z_{i} of  f(*) inside \Gamma and it is given by...

    r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i})\cdot f(z) (4)

    In our case the only pole of f(*) inside \Gamma is z= + j so that is...

    r_{1} = \lim_{ z \rightarrow + j} (z - j)\cdot \frac{e^{j z}}{1+z^{2}} = \frac{e^{-1}}{2 j} (5)

    ... and finally...

    \int_{- \infty}^{+ \infty} \frac {\cos x}{1+x^{2}}\cdot dx = 2 \pi j \cdot r_{1} = \frac{\pi}{e} (6)

    Kind regards

    \chi \sigma
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