Residue Theorem Applied to Calculating ψ(k,t)

[jollage]

I have a doubt on this following procedure using the residue theorem:
Initially we have
ψ(k,t)=$\frac{1}{2\pi}\int_{L_{\omega}}\frac{S(k,\omega)}{D(k,\omega)}e^{-i\omega t}d\omega$
Then the author said using the residue theorem, we have
ψ(k,t)=$-iƩ_{j}\frac{S(k,\omega_j(k))}{\partial D/ \partial \omega (k,\omega_j(k))}e^{-i\omega_j(k) t}$
where $S(k,\omega_j(k)), D(k,\omega_j(k))$ are functions relating $k$ and $\omega$.
What kind of residue theorem is the author using?
Thank you very much

 

[HallsofIvy]

The "residue theorem" the author is using, the only one I know, is
"The integral of f(z) around a closed curve is $2\pi i$ times the sum of the residues at the poles of f inside the curve."

 

[jollage]

The "residue theorem" the author is using, the only one I know, is
"The integral of f(z) around a closed curve is $2\pi i$ times the sum of the residues at the poles of f inside the curve."

Hi,

Thanks for your reply.

I know residue theorem. I just can't relate the usage the author has displayed to the canonical form, like I found in Wiki or what you have said.

To be more specific, could you tell me how did you get the partial derivative of D w.r.t. ω in the denominator?

 

[pasmith]

A pole occurs where $D(k,\omega) = 0$, at which point $\omega = \omega_j(k)$. Locally the Taylor series for D is (for fixed k):
$$D(k,\omega) = (\omega - \omega_j(k))\frac{\partial D}{\partial \omega}(k,\omega_j(k)) + \dots$$
Thus close to $\omega = \omega_j(k)$ (assuming $S(k,\omega_j(k)) \neq 0$) the integrand is approximately
$$\frac{S(k,\omega_j(k))}{(\omega - \omega_j(k))\frac{\partial D}{\partial \omega}(k,\omega_j(k))}e^{-i\omega_j(k)t}$$
so the residue is
$$\frac{S(k,\omega_j(k))}{\frac{\partial D}{\partial \omega}(k,\omega_j(k))}e^{-i\omega_j(k)t}.$$

I don't know where the minus sign in the original comes from; presumably the contour is being traversed clockwise instead of anticlockwise.

 

[Mandelbroth]

Hi,

Thanks for your reply.

I know residue theorem. I just can't relate the usage the author has displayed to the canonical form, like I found in Wiki or what you have said.

To be more specific, could you tell me how did you get the partial derivative of D w.r.t. ω in the denominator?

Suppose the function you are integrating has a simple pole at $z_0$. If the function being integrated can be written in the form $\frac{p(z)}{q(z)}$, with $q(z_0)=0$ and $q'(z_0)\neq 0$, then its residue at the point $z_0$ is given by $\frac{p(z_0)}{q'(z_0)}$.

 

[jollage]

A pole occurs where $D(k,\omega) = 0$, at which point $\omega = \omega_j(k)$. Locally the Taylor series for D is (for fixed k):
$$D(k,\omega) = (\omega - \omega_j(k))\frac{\partial D}{\partial \omega}(k,\omega_j(k)) + \dots$$
Thus close to $\omega = \omega_j(k)$ (assuming $S(k,\omega_j(k)) \neq 0$) the integrand is approximately
$$\frac{S(k,\omega_j(k))}{(\omega - \omega_j(k))\frac{\partial D}{\partial \omega}(k,\omega_j(k))}e^{-i\omega_j(k)t}$$
so the residue is
$$\frac{S(k,\omega_j(k))}{\frac{\partial D}{\partial \omega}(k,\omega_j(k))}e^{-i\omega_j(k)t}.$$

I don't know where the minus sign in the original comes from; presumably the contour is being traversed clockwise instead of anticlockwise.

Thanks a lot, what you said makes sense.