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Math Help - upper half plane

  1. #1
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    upper half plane

    Compute the following integral.

    \int^{\infty}_{0} \frac{x^2}{x^8+1}dx.


    I think that I should get \frac{1}{4} \sqrt{1-\frac{1}{\sqrt 2}}  \cdot \pi. I see we know this is even. Then there is work to show that
    \int^{\infty}_{-\infty} is  2\pi i times the sum of the residues of \frac{z^2}{z^8+1} in the upper half plane. Then I found that

    c_k = \text{exp}[i(\frac{\pi}{8}+\frac{2k\pi}{8}] where k=0,1, \ldots, 7. Then the zeros in the upper half plane occur at k=0,1,2,3.

    Then I used

    \text{Res}_{z=c_k}=\frac{c_k^2}{8c_k^7}=\frac{1}{8  c_k^5}=\frac{1}{8}c_k^{-5} where k=0,1,2,3. Then when I add these residues up I did not get the answer above, I got \frac{\sqrt{\sqrt{2}+2} \cdot \pi}{16} - \frac{\sqrt{2-\sqrt{2}} \cdot \pi }{16}i.
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  2. #2
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    Quote Originally Posted by xboxlive89128 View Post
    Compute the following integral.

    \int^{\infty}_{0} \frac{x^2}{x^8+1}dx.


    I think that I should get \frac{1}{4} \sqrt{1-\frac{1}{\sqrt 2}}  \cdot \pi. I see we know this is even. Then there is work to show that
    \int^{\infty}_{-\infty} is  2\pi i times the sum of the residues of \frac{z^2}{z^8+1} in the upper half plane. Then I found that

    c_k = \text{exp}[i(\frac{\pi}{8}+\frac{2k\pi}{8}] where k=0,1, \ldots, 7. Then the zeros in the upper half plane occur at k=0,1,2,3.

    Then I used

    \text{Res}_{z=c_k}=\frac{c_k^2}{8c_k^7}=\frac{1}{8  c_k^5}=\frac{1}{8}c_k^{-5} where k=0,1,2,3. Then when I add these residues up I did not get the answer above, I got \frac{\sqrt{\sqrt{2}+2} \cdot \pi}{16} - \frac{\sqrt{2-\sqrt{2}} \cdot \pi }{16}i.
    That seems to be pretty much correct, but it's not obvious how to transform your answer to the given one. I would take the residues as e^{ik\pi/8}, where k = 1, 3, 5, 7. Their negative-fifth powers are e^{ik\pi/8}, where k = 1, 7, –3, –5. The sum of the first two of these is 2i\sin(\pi/8), and the sum of the other two is -2i\sin(3\pi/8). So the sum of the four residues is \tfrac i4(\sin(\pi/8) - \sin(3\pi/8)) = -\tfrac i2\cos(\pi/4)\sin(\pi/8) (addition formula for trig functions!). Now use \cos(\pi/4) = 1/\sqrt2 and \sin(\pi/8) = \frac1{\sqrt2}\sqrt{1-\frac1{\sqrt2}} to see that 2\pi i times the sum of the residues is \frac\pi2\sqrt{1-\frac1{\sqrt2}}.

    That is the integral from -\infty to \infty, so you need to divide by 2 to get the integral from 0 to \infty.
    Last edited by Opalg; December 3rd 2009 at 07:50 AM. Reason: typo
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