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Thread: Residue theorem

  1. #1
    Junior Member
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    Residue theorem

    Hi I'm having some problems with the below question

    Use the residue theorem to compute

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


    So i rewrote the intergral as

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

    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...

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

    ... is given by...

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

    ... where $\displaystyle f(z) = \frac{e^{j z}}{1+z^{2}}$ and $\displaystyle \Gamma$ 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...

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

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

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

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

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

    ... and finally...

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

    Kind regards

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