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Math Help - Urgent, residue and winding numbers

  1. #1
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    Exclamation Urgent, residue and winding numbers

    1. Evaluate the integrals

     \int ^{2\pi} _0 \frac{dt}{2+sint}

    and

     \int ^{\infty} _0 \frac{x^2}{x^4 +x}dx

    2. Let z_1,z_2,z_3,z_4 \epsilon \mathbb{C} be the vertices of a closed quadrilateral \Box such that [z_1,z_2] \oplus [z_2,z_3]\oplus [z_3,z_4]\oplus [z_4,z_1] traverses \partial \Box in counterclockwise direction (abusing notation, we shall write \partial \Box for that curve) show that v(\partial \Box,z) is 0 for z\epsilon \mathbb{C} ` \Box (C-minus Box) and 1 if z is an interior point of \Box (Hint: do not attempt to parametrize \partial \Box..)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jbpellerin View Post
    1. Evaluate the integrals

     \int ^{2\pi} _0 \frac{dt}{2+sint}

    and

     \int ^{\infty} _0 \frac{x^2}{x^4 +x}dx

    2. Let z_1,z_2,z_3,z_4 \epsilon \mathbb{C} be the vertices of a closed quadrilateral \Box such that [z_1,z_2] \oplus [z_2,z_3]\oplus [z_3,z_4]\oplus [z_4,z_1] traverses \partial \Box in counterclockwise direction (abusing notation, we shall write \partial \Box for that curve) show that v(\partial \Box,z) is 0 for z\epsilon \mathbb{C} ` \Box (C-minus Box) and 1 if z is an interior point of \Box (Hint: do not attempt to parametrize \partial \Box..)
    For the first ones do you need to do it by CA?
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  3. #3
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    CA? I'm pretty sure the prof wants us to use the residue theorem
    theres an example in the book with cos instead of sin and I'm working on getting it to work for this question
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    Quote Originally Posted by jbpellerin View Post
    1. Evaluate the integrals

     \int ^{2\pi} _0 \frac{dt}{2+sint}
    Comrade.

    Notice that,
    \int_0^{2\pi} \frac{dt}{2+\sin t} = \oint \limits_{|z|=1} \frac{1}{2+\tfrac{1}{2i}(z - z^{-1})} \cdot \frac{1}{iz} dz

    Thus, we get,
     \oint_{|z|=1} \frac{2dz}{z^2+4iz - 1}

    Now apply the residue theorem.
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  5. #5
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    Quote Originally Posted by jbpellerin View Post
     \int ^{\infty} _0 \frac{x^2}{x^4 +x}dx
    By integrating over partial circular contour of angle 2\pi/n we get \int_0^{\infty} \frac{dx}{x^n+1} = \frac{(\pi/n)}{\sin (\pi/n)} for n\geq 2.

    \int_0^{\infty} \frac{x^2}{x^4+x} dx=\int_0^{\infty} \frac{x}{x^3+1} = \int_0^{\infty} \frac{dx}{x^2-x+1} - \int_0^{\infty} \frac{dx}{x^3+1}

    Now,
    \int_0^{\infty} \frac{dx}{x^2-x+1} = \int_0^{\infty} \frac{dx}{\left( x - \frac{1}{2} \right)^2 + \frac{3}{4} } = \int_{-1/2}^{\infty} \frac{dt}{t^2 + \frac{3}{4}}
    That computes to,
    \frac{2}{\sqrt{3}}\tan^{-1} \left( \frac{2t}{\sqrt{3}} \right) \bigg|_{-1/2}^{\infty} = \frac{\pi}{\sqrt{3}} + \frac{\pi}{3\sqrt{3}} = \frac{4\pi}{3\sqrt{3}}

    And by above result,
    \int_0^{\infty}\frac{dx}{x^3+1} = \frac{(\pi/3)}{\sin (\pi/3)} = \frac{2\pi}{3\sqrt{3}}

    Thus, the final answer is, \frac{4\pi}{3\sqrt{3}}-\frac{2\pi}{3\sqrt{3}} = \frac{2\pi}{3\sqrt{3}}
    Last edited by ThePerfectHacker; November 12th 2008 at 06:07 AM.
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  6. #6
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    Quote Originally Posted by jbpellerin View Post
    2. Let z_1,z_2,z_3,z_4 \epsilon \mathbb{C} be the vertices of a closed quadrilateral \Box such that [z_1,z_2] \oplus [z_2,z_3]\oplus [z_3,z_4]\oplus [z_4,z_1] traverses \partial \Box in counterclockwise direction (abusing notation, we shall write \partial \Box for that curve) show that v(\partial \Box,z) is 0 for z\epsilon \mathbb{C} ` \Box (C-minus Box) and 1 if z is an interior point of \Box (Hint: do not attempt to parametrize \partial \Box..)
    If z is outside the rectange then computing the winding number is to evaluate \oint_{\partial \Box}\frac{dw}{w-z}=0 by Cauchy's theorem. If z is inside then it can be shown, here (post #8 with long proof), that if C is a tiny circle around z still inside \partial \Box then \oint_{\partial \Box} \frac{dw}{w-z} = \oint_{C} \frac{dw}{w-z} = 2\pi i. And so the winding number is 1.
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  7. #7
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    Whenever I make a thread on here I always hope you're online because you always have excellent answers
    thank you again for the help
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  8. #8
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    Quote Originally Posted by jbpellerin View Post
    Whenever I make a thread on here I always hope you're online because you always have excellent answers
    You should not say that because I have a terrible ego problem.
    .....Many members here compain about my addiction of myself to myself.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    .....Many members here compain about my addiction of myself to myself.
    Here here! ...jk
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