How can I use the Greens function method to solve a PDE with the given equation?

[hunt_mat]

I have to solve the following PDE:
<br /> \frac{\partial^{2}u}{\partial t^{2}}+2\frac{\partial^{2}u}{\partial t\partial x}+\frac{\partial u}{\partial x}+\frac{\partial u}{\partial t}+k^{2}u=f<br />
I use the Greens function method and examine the equation:
<br /> \frac{\partial^{2}G}{\partial t^{2}}+2\frac{\partial^{2}G}{\partial t\partial x}+\frac{\partial G}{\partial x}+\frac{\partial G}{\partial t}+k^{2}G=4\pi\delta (x-x&#039;)\delta (t-t&#039;)<br />
I then write:
<br /> G=\frac{1}{2\pi}\int_{-\infty}^{\infty}g(X|X&#039;)e^{i\omega (T-T&#039;)}d\omega\quad \delta (T-T&#039;) =\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega (T-T&#039;)}d\omega<br />
The equation then becomes:
<br /> (1+2\omega i)\frac{\partial g}{\partial X}+(k^{2}-\omega^{2}+i\omega)g=4\pi\delta (X-X&#039;) <br />
Take the Fourier transform to obtain:
<br /> i\xi (1+2\omega i)\hat{g}+(k^{2}-\omega^{2}-\omega i)\hat{g}=4\pi e^{i\xi X&#039;}<br />
Rearrange and take the inverse Fourier transform to obtain:
<br /> g=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{4\pi e^{-i\xi (X-X&#039;)}}{i\xi (1+2\omega i)+k^{2}-\omega^{2}-\omega i}d\xi<br />
Am I on the right track here?

 

[hunt_mat]

I think that I can find g by using contour integration. Write:
<br /> \frac{1}{2\pi}\oint_{\gamma}\frac{4\pi e^{-iz(X-X&#039;)}}{iz(1+2\omega i)+k^{2}-\omega^{2}-\omega i}dz<br />
Which can then be evaluated via Cauchy's integral formula:
<br /> g=4\pi ie^{i(X-X&#039;)h(\omega )},\quad h(\omega )=\frac{k^{2}-\omega^{2}-\omega i}{1+2\omega i}<br />

Thoughts?