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Math Help - another complex question

  1. #1
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    another complex question

    \int_{-\infty}^{\infty}\frac{1}{x^{6}+1}dx

    =\int_{C}\frac{1}{z^{6}+1}dz i think.... a bit confused because there will be singularities in UHP and LHP...

    =\int_{C}\frac{1}{(z^{2}+1)(z^{4}-z^{2}+1)}dz

    =\int_{C}\frac{1}{(z-i)(z+i)(z^{4}-z^{2}+1)}dz

    let z^4=z^2

    z^{2}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{3}\\i}{2}

    \\\\z=\{-i,i,\frac{1}{2}(\sqrt{3}+i),\frac{1}{2}(\sqrt{3}-i),\frac{1}{2}(-\sqrt{3}+i),\frac{1}{2}(-\sqrt{3}-i)\}

    now from here... how the hell do i do the residues.. the singularities are all over the place???

    please someone help
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fredrick08 View Post
    \int_{-\infty}^{\infty}\frac{1}{x^{6}+1}dx

    =\int_{C}\frac{1}{z^{6}+1}dz i think.... a bit confused because there will be singularities in UHP and LHP...

    =\int_{C}\frac{1}{(z^{2}+1)(z^{4}-z^{2}+1)}dz

    =\int_{C}\frac{1}{(z-i)(z+i)(z^{4}-z^{2}+1)}dz

    let z^4=z^2

    z^{2}=\frac{1\pm\sqrt{1-4}}{2}=\frac{1\pm\sqrt{3}\\i}{2}

    \\\\z=\{-i,i,\frac{1}{2}(\sqrt{3}+i),\frac{1}{2}(\sqrt{3}-i),\frac{1}{2}(-\sqrt{3}+i),\frac{1}{2}(-\sqrt{3}-i)\}

    now from here... how the hell do i do the residues.. the singularities are all over the place???

    please someone help
    What do you mean how do you find the residues?? How else does one usually find them?
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  3. #3
    MHF Contributor chisigma's Avatar
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    The poles of the function f(z)= \frac{1}{1+z^{6}} are at z= e^{\pm i \pi \frac{1+2k}{6}} , k=0,1,2. Only the residues of the poles with positive imaginary part give contribution of the integral that is...

    \int_{-\infty}^{+\infty} \frac{dx}{1+x^{6}} = 2 \pi i \sum_{k=0}^{2} r_{k} (1)

    ... where...

    r_{0} = \lim_{z \rightarrow e^{i \frac{\pi}{6}}} \frac{z-e^{i \frac{\pi}{6}}}{1+z^{6}}

    r_{1} = \lim_{z \rightarrow e^{i \frac{\pi}{2}}} \frac{z-e^{i \frac{\pi}{2}}}{1+z^{6}}

    r_{2} = \lim_{z \rightarrow e^{i \frac{5 \pi}{6}}} \frac{z-e^{i \frac{5 \pi}{6}}}{1+z^{6}} (2)

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    The poles of the function f(z)= \frac{1}{1+z^{6}} are at z= e^{\pm i \pi \frac{1+2k}{6}} , k=0,1,2. Only the residues of the poles with positive imaginary part give contribution of the integral that is...

    \int_{-\infty}^{+\infty} \frac{dx}{1+x^{6}} = 2 \pi i \sum_{k=0}^{2} r_{k} (1)

    ... where...

    r_{0} = \lim_{z \rightarrow e^{i \frac{\pi}{6}}} \frac{z-e^{i \frac{\pi}{6}}}{1+z^{6}}

    r_{1} = \lim_{z \rightarrow e^{i \frac{\pi}{2}}} \frac{z-e^{i \frac{\pi}{2}}}{1+z^{6}}

    r_{2} = \lim_{z \rightarrow e^{i \frac{5 \pi}{6}}} \frac{z-e^{i \frac{5 \pi}{6}}}{1+z^{6}} (2)

    Kind regards

    \chi \sigma
    srry im having a lot of trouble with these limits?? dont they all goto zero.. as z approaches to what equal in the top line????

    srry dumb question.... Thankyou very much for ur help
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  5. #5
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    No, the limits are not all 0. If z= e^{i\pi/6} the numerator is 0 but z^6= e^{i\pi}= -1 so the denominator, z^6+ 1 is also 0. Use L'Hopital's rule to find the limits.
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  6. #6
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    I think there's a simpler way. Consider

    f(z)=\frac{p(z)}{q(z)} where

    p(z) = 1 and

    q(z)=z^6+1

    Then p(z_i) \neq 0, q(z_i)=0 and q'(z_i) \neq 0 for each of your singularities.

    Then the residue at z_i is \frac{p(z_i)}{q'(z_i)}=\frac{1}{6{z_i}^5}

    There are usually singularities in the UHP and LHP with these problems, but the LHP singularities are never inside the contour that you construct.
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