Changes

Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.

Now, we have an equation with integrals on both sides. Since both of these integrals have the same limits and integrands, they must be integrals of equal functions. We can simply drop the integrals.

<math> \frac{d^2 \frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}}{dt^2}-c^2\frac{d^2\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}}{dx^2} = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>

<math> \frac{d^2}{dt^2} \left(\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right)-c^2\frac{d^2}{dx^2}\left(\frac{1}{4\pi^2} G(\Delta \mathbf{x},\Delta t)e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}\right) = \frac{1}{16\pi^4} e^{-i\mathbf{k}\cdot\Delta \mathbf{x}}e^{i\omega \Delta t}</math>

This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into

This is a complicated equation, but it can be solved for G. Once G is calculated, we can apply an inverse Fourier transform and find g; we can then plug this into