Results 1 to 4 of 4

Thread: Complex Analysis

  1. #1
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411

    Complex Analysis

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

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

    For $\displaystyle n\geq 2$ this is easy to show. But how to show this for n=1?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Hmm.

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

    Thus,

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

    The RHS goes to zero as $\displaystyle R\to\infty.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    I don't see how this explains $\displaystyle \int_{\gamma_R}\frac{dz}{z^2+1}\to 0$

    We can only state this with the knowledge that:

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

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

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

    But for n=1, we can't tell this easy....at least i don't see how. Maybe with a different $\displaystyle \gamma_R$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    Well, I think the approach works fine. But if you don't like it, you could also use an ML estimate. You have $\displaystyle L=\pi R,$ and

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

    $\displaystyle \displaystyle\left|\int_{\gamma_{R}}\frac{1}{z^{2} +1}\,dz\right|\le ML=\frac{\pi R}{R^{2}}=\frac{\pi}{R}\to 0$ as $\displaystyle R\to\infty.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 4th 2011, 05:30 AM
  2. Replies: 6
    Last Post: Sep 13th 2011, 07:16 AM
  3. Replies: 1
    Last Post: Oct 2nd 2010, 01:54 PM
  4. Replies: 12
    Last Post: Jun 2nd 2010, 02:30 PM
  5. Replies: 1
    Last Post: Mar 3rd 2008, 07:17 AM

Search Tags


/mathhelpforum @mathhelpforum