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Thread: cauchy integral formula

  1. #1
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    cauchy integral formula

    Eval

    $\displaystyle \frac{1}{2\pi i}\oint\frac{e^{tz}}{(z^2+1)^2}dz$ if t>0 and C is the circle {|z|=3}

    Hmm...

    Am I suppose to parameterise something?

    I can write for example z(t)=3e^{it} then dz = 3i e^{it}dt and sub it back in, but that just makes it messy and complicated.

    Am I missing an easier way?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Eval

    $\displaystyle \frac{1}{2\pi i}\oint\frac{e^{tz}}{(z^2+1)^2}dz$ if t>0 and C is the circle {|z|=3}

    Hmm...

    Am I suppose to parameterise something?

    I can write for example z(t)=3e^{it} then dz = 3i e^{it}dt and sub it back in, but that just makes it messy and complicated.

    Am I missing an easier way?
    Do you know the Residue theorem? You're function has singularities at $\displaystyle z=\pm i$ both of which lie in $\displaystyle C$. So, $\displaystyle \oint_{C}f(z)dz=\underset{z= i}{\text{Res}}f(z)+\underset{z=- i}{\text{Res}}f(z)$
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  3. #3
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    Thanks, I considered that just now, but this question belongs to a section well-prior to covering the residue thm, in the CIF section. So I suppose it should be solvable using that...
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    Thanks, I considered that just now, but this question belongs to a section well-prior to covering the residue thm, in the CIF section. So I suppose it should be solvable using that...
    Ok, decompose your the denominator into $\displaystyle (z-i)^2(z+i)^2$ and decompose this using partial fractions and then from there apply Cauchy's integral theorem.
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  5. #5
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    I get

    $\displaystyle -\frac{t}{2} \cos t$

    but the answer says: $\displaystyle 1/2 (\sin t -t \cos t)$

    for the partial fractions I got A = B = -1/4 e^{tz}

    so we have

    $\displaystyle \frac{1}{2\pi i}
    \oint_C\frac{e^{tz}}{(z^2+1)^2}dz=
    $

    $\displaystyle \frac{1}{2\pi i}
    \left\{-\left(\oint_C\frac{e^{tz}/4}{(z+i)^2}\; dz +
    \oint_C\frac{e^{tz}/4}{(z-i)^2}\;dz\right)\right\}=$

    $\displaystyle -\frac{1}{2\pi i}2\pi i \frac{1}{4}(f'(-i)+f'(i))$ where

    $\displaystyle f(z)=e^{tz}$ and $\displaystyle f'(z)=te^{tz}$

    $\displaystyle =-\frac{t}{4}(e^{-it}+e^{it})$

    $\displaystyle =-\frac{t}{4}(2\cos t)$

    $\displaystyle =-\frac{t}{2}\cos t$

    Did I make a mistake?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by scorpion007 View Post
    I get

    $\displaystyle -\frac{t}{2} \cos t$

    but the answer says: $\displaystyle 1/2 (\sin t -t \cos t)$

    for the partial fractions I got A = B = -1/4 e^{tz}

    so we have

    $\displaystyle \frac{1}{2\pi i}
    \oint_C\frac{e^{tz}}{(z^2+1)^2}dz=
    $

    $\displaystyle \frac{1}{2\pi i}
    \left\{-\left(\oint_C\frac{e^{tz}/4}{(z+i)^2}\; dz +
    \oint_C\frac{e^{tz}/4}{(z-i)^2}\;dz\right)\right\}=$

    $\displaystyle -\frac{1}{2\pi i}2\pi i \frac{1}{4}(f'(-i)+f'(i))$ where

    $\displaystyle f(z)=e^{tz}$ and $\displaystyle f'(z)=te^{tz}$

    $\displaystyle =-\frac{t}{4}(e^{-it}+e^{it})$

    $\displaystyle =-\frac{t}{4}(2\cos t)$

    $\displaystyle =-\frac{t}{2}\cos t$

    Did I make a mistake?
    I don't think you decomposed it correct.
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  7. #7
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    ah yes! thanks!
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