Does this have to do with the orientation?

747
2
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
 
747
2
Re: Integral

Does this have to do with the orientation?
 

Dick

Science Advisor
Homework Helper
26,249
611
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.
 
747
2
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.
 

Dick

Science Advisor
Homework Helper
26,249
611
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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