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Math Help - Complex Analysis

  1. #1
    Senior Member Dinkydoe's Avatar
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    Complex Analysis

    I want to show that the integral \int_{\gamma_R}\frac{dz}{z^{2n}+1} \to  0<br />
as R\to \infty

    ,where \gamma_R is the upper semi-circle from R to -R

    For n\geq 2 this is easy to show. But how to show this for n=1?
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  2. #2
    A Plied Mathematician
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    Hmm.

    \displaystyle\int_{\gamma_{R}}\frac{1}{z^{2}+1}\,d  z+\int_{-R}^{R}\frac{1}{x^{2}+1}\,dx=2\pi i\text{Res}[\frac{1}{z^{2}+1},i], right?

    Thus,

    \displaystyle\int_{\gamma_{R}}\frac{1}{z^{2}+1}\,d  z=2\pi i\text{Res}[\frac{1}{z^{2}+1},i]-\int_{-R}^{R}\frac{1}{x^{2}+1}\,dx.

    The RHS goes to zero as R\to\infty.
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  3. #3
    Senior Member Dinkydoe's Avatar
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    I don't see how this explains \int_{\gamma_R}\frac{dz}{z^2+1}\to 0

    We can only state this with the knowledge that:

    \int_{-\infty}^{\infty}\frac{1}{x^2+1}dx = 2\pi i\mathrm{Res}[\frac{1}{z^2+1},i], wich frankly we don't.

    The way it can be shown for n\geq 2 is by using \gamma_R(t)=Re^{2\pi i t},  t\in [0,1]

    Then \int_{\gamma_R}\frac{dz}{1+z^{2n}} \leq \sup_{t\in [0,1]}|\frac{2\pi iRe^{2\pi i t}}{(Re^{2\pi i t})^{2n}+1}|\cdot \pi R\to 0 as R\to\infty

    But for n=1, we can't tell this easy....at least i don't see how. Maybe with a different \gamma_R
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  4. #4
    A Plied Mathematician
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    Well, I think the approach works fine. But if you don't like it, you could also use an ML estimate. You have L=\pi R, and

    \left|\dfrac{1}{R^{2}+1}\right|\le\left|\dfrac{1}{  R^{2}}\right|=:M. Therefore,

    \displaystyle\left|\int_{\gamma_{R}}\frac{1}{z^{2}  +1}\,dz\right|\le ML=\frac{\pi R}{R^{2}}=\frac{\pi}{R}\to 0 as R\to\infty.
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