Fourier transform of Green Function

Question

To work on a electrostatic potential, we need to use fourier transformation of the equation,

$$\nabla^2 G (R) = \delta (R)$$

We get the equation:

$$G(R) = \frac{1}{({2 \pi})^3} \int d^3 k \ \ \ e^{i \vec{k} \cdot \vec{R}} G( \tilde{R})$$

I know that the cube over $2 \pi$ due to the three dimensional space but why do we get the $ G( \tilde{R})$ in the right side of the Fourier transformation?

Answer 1

It works roughly like this:

You Fourier-transform both sides of the initial equation $\Delta G(R) = \delta(R)$. The differential operators in $\Delta = \sum \partial_i^2$ should transform into some multiplicative operators $\sum k_i^2$. You can then invert algebraically to get $G(R)$ and finally back-transform that expression.