How to compute the Fourier transform of an ellipse (the interior), please?
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$F(u, v) = \mathcal{F} \left\{ f(x, y) \right\} = \int\int f(x, y) e^{2\pi i (ux + vy)}\, d x d y =\int \left[ \int f(x,y) e^{2\pi i ux}d x \right] e^{2\pi i v y }d y $$
I assume that I need an ellipse in the polar coordinate system, right?
For a circle, it is like this: Yes, I mean with its interior. For a circle it is like this: https://bpb-us-w2.wpmucdn.com/sites.coecis.cornell.edu/dist/c/442/files/2021/05/Fourier-Transforms-2D.pdf
$$ \iint_{x^2/a^2+y^2/b^2\le 1} e^{2\pi i ux}e^{2\pi i vy} dxdy $$ $$ =\int_{-b}^b dy e^{2\pi i vy} \int_{-a\sqrt{1-y^2/b^2}}^{a\sqrt{1-y^2/b^2}} e^{2\pi i ux}dx $$ The substitution $x'=x/a$, $y'=y/b$, then renaming $y'\to y$, $x'\to x$ yields $$ =ab\int_{-1}^1 dy e^{2\pi i vby} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} e^{2\pi i uax}dx $$ $$ = ab \frac{1}{2\pi i u a} \int_{-1}^1 dy e^{2\pi i vby} e^{2\pi i u a x}\mid _{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} $$ $$ = b \frac{1}{2\pi i u} \int_{-1}^1 dy e^{2\pi i vby} [e^{2\pi i u a \sqrt{1-y^2}} - e^{-2\pi i u a \sqrt{1-y^2}} ] $$ The substitution $y=\sin\alpha$, $dy=\cos\alpha d\alpha$ yields $$ = b \frac{1}{2\pi iu} \int_{-\pi/2}^{\pi/2} d\alpha e^{2\pi i vb\sin\alpha} [e^{2\pi i u a \cos\alpha} - e^{-2\pi i u a \cos\alpha} ]\cos\alpha $$ Euler's formuals for the complex exponential $(e^{ix}-e^{-ix})/(2i)$ yield $$ = b \frac{1}{\pi u} \int_{-\pi/2}^{\pi/2} d\alpha e^{2\pi i vb\sin\alpha} \sin(2\pi u a \cos\alpha) \cos\alpha $$ $$ = b \frac{1}{\pi u} \int_0^{\pi/2} d\alpha e^{2\pi i vb\sin\alpha} \sin(2\pi u a \cos\alpha) \cos\alpha +b \frac{1}{\pi u} \int_{-\pi/2}^0 d\alpha e^{2\pi i vb\sin\alpha} \sin(2\pi u a \cos\alpha) \cos\alpha $$ The substitution $\alpha =-\tau$ in the second of these integrals, $\cos(-\tau)=\cos(\tau)$ and $\sin(-\tau)=-\sin\tau$ plus Euler's formula for $(e^{ix}+e^{-ix})/2$ yields $$ = 2b \frac{1}{\pi u} \int_0^{\pi/2} d\alpha \cos(2\pi vb\sin\alpha) \sin(2\pi u a \cos\alpha) \cos\alpha $$
$$ = 2b \frac{1}{\pi u} \int_0^{\pi/2} d\alpha \sin(2\pi u a \cos\alpha) \cos(2\pi vb\sin\alpha) \cos\alpha $$ $$ = b \frac{1}{\pi u} \int_0^{\pi/2} d\alpha [\sin(2\pi (ua\cos\alpha -vb\sin\alpha) ) +\sin(2\pi (ua\cos\alpha+vb\sin\alpha) ) ] \cos\alpha $$ Define the unitless amplitude $c\equiv \sqrt{(ua)^2+(vb)^2}$, $\cos \beta=ua/c$, $\sin\beta=vb/c$, $$ = b \frac{1}{\pi u} \int_0^{\pi/2} d\alpha [\sin(2\pi c (\cos\alpha \cos\beta -\sin\alpha\sin\beta) ) +\sin(2\pi c (\cos\alpha\cos\beta+\sin\alpha\sin\beta) ) ] \cos\alpha $$ $$ = b \frac{1}{\pi u} \int_0^{\pi/2} d\alpha [\sin(2\pi c \cos(\alpha +\beta)) +\sin(2\pi c \cos(\alpha-\beta)) ] \cos\alpha $$ $$ = -b \frac{1}{\pi u} \int_0^{-\pi/2} d\alpha [\sin(2\pi c \cos(-\alpha +\beta)) +\sin(2\pi c \cos(-\alpha-\beta)) ] \cos(-\alpha) $$ $$ = b \frac{1}{\pi u} \int_{-\pi/2}^0 d\alpha [\sin(2\pi c \cos(\alpha +\beta)) +\sin(2\pi c \cos(\alpha-\beta)) ] \cos(\alpha) $$ $$ = b \frac{1}{2\pi u} \int_{-\pi/2}^{\pi/2} d\alpha [\sin(2\pi c \cos(\alpha +\beta)) +\sin(2\pi c \cos(\alpha-\beta)) ] \cos(\alpha) $$ $$ = b \frac{1}{2\pi u} \int_{\pi/2}^{3\pi/2} d\alpha [\sin(2\pi c \cos(\alpha -\pi +\beta)) +\sin(2\pi c \cos(\alpha-\pi -\beta)) ] \cos(\alpha-\pi) $$ $$ = b \frac{1}{2\pi u} \int_{\pi/2}^{3\pi/2} d\alpha [-\sin(2\pi c \cos(\alpha +\beta)) -\sin(2\pi c \cos(\alpha -\beta)) ] (-)\cos(\alpha) $$ $$ = b \frac{1}{4\pi u} \int_{-\pi/2}^{3\pi/2} d\alpha [\sin(2\pi c \cos(\alpha +\beta)) +\sin(2\pi c \cos(\alpha -\beta)) ] \cos(\alpha) $$
$$ = b \frac{1}{4\pi u} \int_{-\pi/2}^{3\pi/2} d\alpha \sin(2\pi c \cos(\alpha +\beta)) \cos\alpha + b \frac{1}{4\pi u} \int_{-\pi/2}^{3\pi/2} d\alpha \sin(2\pi c \cos(\alpha-\beta)) \cos\alpha $$
$$ = b \frac{1}{4\pi u} \int_{-\pi/2+\beta}^{3\pi/2+\beta} d\alpha \sin(2\pi c \cos \alpha ) \cos(\alpha-\beta) + b \frac{1}{4\pi u} \int_{-\pi/2-\beta}^{3\pi/2-\beta} d\alpha \sin(2\pi c \cos \alpha ) \cos(\alpha+\beta) $$
$$ = b \frac{1}{4\pi u} \int_{-\pi/2+\beta}^{3\pi/2+\beta} d\alpha \sin(2\pi c \cos \alpha ) \cos \alpha \cos\beta +b \frac{1}{4\pi u} \int_{-\pi/2+\beta}^{3\pi/2+\beta} d\alpha \sin(2\pi c \cos \alpha ) \sin \alpha \sin\beta + (\beta \leftrightarrow -\beta) $$
$$ = b \frac{1}{4\pi u} \cos\beta \int_0^{2\pi} d\alpha \sin(2\pi c \cos \alpha ) \cos \alpha +b \frac{1}{4\pi u} \sin\beta \int_0^{2\pi} d\alpha \sin(2\pi c \cos \alpha ) \sin \alpha + (\beta \leftrightarrow -\beta) $$
$$ = b \frac{1}{4\pi u} \cos\beta \int_0^{2\pi} d\alpha \sin(2\pi c \cos \alpha ) \cos \alpha + (\beta \leftrightarrow -\beta) $$ $$ = b \frac{1}{2\pi u} \cos\beta \int_0^{\pi} d\alpha \sin(2\pi c \cos \alpha ) \cos \alpha + (\beta \leftrightarrow -\beta) $$
$$ = b \frac{1}{2 u} \cos\beta J_1(2\pi c) + (\beta \leftrightarrow -\beta) $$ $$ = b \frac{1}{u} \cos\beta J_1(2\pi c) = \frac{ab}{c} J_1(2\pi c) $$