1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook