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Homework Help: Does this have to do with the orientation?

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The final answer should be
    [tex]
    \begin{equation}
    + g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} = + g^{2} \frac{1}{s+\gamma+\gamma_{n}}
    \end{equation}
    [/tex]

    but I get a - in front of g

    [tex]
    \begin{subequations}
    \begin{eqnarray}
    \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'}
    &=&
    \frac{1}{\pi} \int_{-\infty}^{\infty} d \Delta'\; \frac{\gamma_{n}}{(\Delta'^{2}+\gamma_{n}^{2})}\, \frac{1}{(s+\gamma+i\Delta')}
    \\
    \nonumber
    &=&
    \frac{1}{\pi} \int_{-\infty}^{\infty} d \Delta'\; \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')}
    \end{eqnarray}
    \end{subequations}
    [/tex]
    in which case the poles are at $\pm i\gamma_{n}, +i(s+\gamma)$.

    [tex]
    \begin{subequations}
    \begin{eqnarray}
    \lim \limits_{\Delta' \to i\gamma_{n}} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')} &=& \frac{1}{2i}\frac{1}{s+\gamma-\gamma_{n}}
    \\
    \lim \limits_{\Delta' \to -i\gamma_{n}} \frac{\gamma_{n}}{(\Delta'-i\gamma_{n})}\, \frac{1}{(s+\gamma+i\Delta')} &=& \frac{-1}{2i}\frac{1}{s+\gamma+\gamma_{n}}
    \\
    \lim \limits_{\Delta' \to i(s+\gamma)} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})}\frac{1}{(s+\gamma+i\Delta')} &=& \lim \limits_{\Delta' \to i(s+\gamma)} \frac{\gamma_{n}}{(\Delta'+i\gamma_{n})(\Delta'-i\gamma_{n})}\frac{-i}{(-i(s+\gamma)+\Delta')}
    \\
    \end{eqnarray}
    \end{subequations}
    [/tex]
    It is simpler integrate in down region where we have only one pole [tex] $ -i\gamma_{n}$.[/tex]
    While we have two poles in the region above the x:
    [tex]$\Delta'= i\gamma_n$,[/tex]
    and
    [tex] $\Delta'= i(s+\gamma)$.[/tex]
    [tex]
    \begin{subequations}
    \begin{eqnarray}
    \oint_{\gamma_{R}(t)} f(z) \, dz &=& 2\pi i[\frac{1}{2i}\frac{-1}{s+\gamma+\gamma_{n}} ]
    %\frac{-1}{2i}\frac{1}{s+\gamma-\gamma_{n}} + \frac{\gamma_{n}}{(-s-\gamma+\gamma_{n})(s+\gamma+\gamma_{n})}]
    \\
    %&=&2\pi i[-\frac{(1+i)\gamma_{n}}{(s+\gamma-\gamma_{n})(s+\gamma+\gamma_{n})}]
    \end{eqnarray}
    \end{subequations}
    [/tex]
    [tex]
    \begin{subequations}
    \begin{eqnarray}
    \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'}
    &=&
    \frac{-\pi}{s+\gamma+\gamma_{n}}
    \\
    \frac{1}{\pi}\int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'}
    &=&
    \frac{-1}{s+\gamma+\gamma_{n}}
    \end{eqnarray}
    \end{subequations}
    [/tex]
    I get:
    [tex]
    \begin{equation}
    + g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} = s + \kappa - g^{2} \frac{1}{s+\gamma+\gamma_{n}}
    \end{equation}
    [/tex]
    2. Relevant equations

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 21, 2009 #2
    Re: Integral

    Does this have to do with the orientation?
     
  4. Feb 25, 2009 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Integral

    I didn't check the whole thing, but if you are integrating in the lower half plane, the contour does close clockwise instead of counterclockwise, so there is an extra factor of -1.
     
  5. Feb 25, 2009 #4
    Re: Integral

    If I'm integrating in the bottom region, how do I know if I'm going cw or ccw? Sorry it's been along time since I've done this.
     
  6. Feb 25, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Re: Integral

    S'ok. You have go in a positive direction (since that's the direction of the real integral) along the x-axis and then circle around the singularity in the negative half plane and then return to the x-axis. You are sort of stuck moving clockwise.
     
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