# Does this have to do with the orientation?

#### Nusc

1. The problem statement, all variables and given/known data
$$+ g^{2} \int_{-\infty}^{\infty} d \Delta'\; {\cal \rho}(\Delta')\, \frac{1}{s+\gamma+i\Delta'} = + g^{2} \frac{1}{s+\gamma+\gamma_{n}}$$

but I get a - in front of g

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

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

3. The attempt at a solution

#### Nusc

Re: Integral

Does this have to do with the orientation?

#### Dick

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.

#### Nusc

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

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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