I'm asked to compute the two-dimensional Fourier Transform of the following function:
$$z = sin(ax+by)$$ where $a,b \in \mathbb{R}$.
What I've tried is the following:
$$ \mathcal{FT} \{ sin(ax+by) \} = \int_{x \in X} \int_{ y \in Y} sin(ax+by)e^{-j2\pi f_x x} e^{-j2\pi f_y y} dx dy = \iint_{x,y} sin(ax+by)e^{-j2\pi f_x x -j2\pi f_y y} dx dy $$
Because $e^ae^b=e^{a+b}$, we can merge the exponentials.
Since the sine can be separated as a sum of exponentials, i.e. $sin(x) = \frac{e^{jx}-e^{-jx}}{2j}$ and let us assume that $c = ax + by$; then:
$$... = \frac{1}{2j}\iint_{x,y} (e^{jc}-e^{-jc}) e^{-j2\pi f_x x -j2\pi f_y y} dx dy = \frac{1}{2j}\iint_{x,y} (e^{jc} e^{-j2\pi f_x x -j2\pi f_y y} - e^{-jc}e^{-j2\pi f_x x -j2\pi f_y y})dxdy =$$
$$= \frac{1}{2j}\iint_{x,y} (e^{jc -j2\pi f_x x -j2\pi f_y y} - e^{-jc -j2\pi f_x x -j2\pi f_y y})dxdy = \frac{1}{2j}\iint_{x,y} (e^{jc -j2\pi f_x x -j2\pi f_y y})dxdy - \frac{1}{2j}\iint_{x,y} e^{-jc -j2\pi f_x x -j2\pi f_y y})dxdy$$
Since a sum in a integral can be split into a sum of two integrals, as shown in the last step.
Now, I'm going to focus only on the first term, i.e. $\frac{1}{2j}\iint_{x,y} (e^{jc -j2\pi f_x x -j2\pi f_y y})dxdy$. First, let me develop the exponential with $c = ax + by$:
$$jc -j2\pi f_x x -j2\pi f_y y = j(ax + by) -j2\pi f_x x -j2\pi f_y y = $$
$$ = jax + jby - j2\pi f_x x -j2\pi f_y y = jx(a - 2 \pi f_x) + jy(b - 2 \pi f_y) = j2\pi x (\frac{a}{2\pi} - f_x) + j2\pi y (\frac{b}{2\pi} - f_y) $$
Therefore, we can now split the exponentials in the following manner:
$$ \frac{1}{2j}\iint_{x,y} e^{j2\pi x (\frac{a}{2\pi} - f_x) + j2\pi y (\frac{b}{2\pi} - f_y)}dxdy = \frac{1}{2j} \iint_{x,y} e^{j2\pi x (\frac{a}{2\pi} - f_x)} e^{j2\pi y (\frac{b}{2\pi} - f_y)}dxdy = \frac{1}{2j} \int_{x} e^{j2\pi x (\frac{a}{2\pi} - f_x)} dx \int_{y} e^{j2\pi y (\frac{b}{2\pi} - f_y)}dy$$
And given that we know that $ \mathcal{FT}\{ 1 \} = \int_{x} e^{-j 2 \pi f_0 x} dx = \delta (f_0)$, then:
$$ ... = \frac{1}{2j} \int_{x} e^{j2\pi x (\frac{a}{2\pi} - f_x)} dx \int_{y} e^{j2\pi y (\frac{b}{2\pi} - f_y)}dy = \frac{1}{2j} \delta ( f_x - \frac{a}{2\pi}) \delta ( f_y - \frac{b}{2\pi})$$
And I suppose that now I have to do the same thing for the second term. Am I in the right direction?