Fourier Transform and Inverse For $x^2>c^2;g(x)=0$ while for $-c\le x\le c;g(x)=\sqrt{c^2-x^2}$
Find the Fourier Transform of g(x), what integral is produced from the inverse transform?  I have calculated the fourier transform as $$F(w)=\frac{\sqrt{\frac{\pi}{2}}\space c J_1(cw)}{w}$$Where J represents a bessel function of the first kind. I am unsure what the inverse transform produces as I get a function which seems to represent a circle? I'm at a loss on how to proceed on that portion of this problem.
 A: Let $F(\omega)$ be given by the integral 
$$\begin{align}
F(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \left(\sqrt{\frac{\pi}{2}}\frac{cJ_1(c\omega)}{\omega}\right)\,e^{i\omega x}\,d\omega \\\\
&=\frac c2 \int_{-\infty}^\infty \frac{J_1(c\omega )}{\omega}\,e^{i\omega x}\,d\omega \\\\
&=\frac c2 \int_{-\infty}^\infty \frac{J_1(\omega )}{\omega}\,e^{i\omega x/c}\,d\omega \tag1
\end{align}$$
In THIS ANSWER, I showed from the integral representation of the $J_0$ Bessel function, $\displaystyle J_0(\omega)=\frac1{2\pi}\int_{-\pi}^\pi e^{i\omega \sin(\phi)}\,d\phi$, that $J_0(\omega)$ can be written 
$$J_0(\omega)=\frac1\pi\int_{-1}^1 \frac{e^{ik\omega}}{\sqrt{1-k^2}}\,dk \tag2$$
Differentiating $(2)$ and using $J_0'=-J_1$ reveals 
$$J_1(\omega)=-\frac1\pi\int_{-1}^1 \frac{ike^{ik\omega}}{\sqrt{1-k^2}}\,dk \tag3$$
Using $(3)$ in $(1)$, we obtain 
$$\begin{align}
F(\omega)&=-\frac{ic}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega x/c}}{\omega}\,\int_{-1}^1\frac{k}{\sqrt{1-k^2}}e^{ik\omega}\,dk\,d\omega\\\\
&=\frac{c}{2\pi}\int_{-1}^1 \frac{k}{\sqrt{1-k^2}}\int_{-\infty}^\infty \frac{\sin(\omega(k+x/c))}{\omega}\,d\omega\,dk\\\\
&=\frac{c}{2}\int_{-1}^1 \frac{k}{\sqrt{1-k^2}}\,\text{sgn}(k+x/c)\,dk\tag 4
\end{align}$$
The integral in $(4)$ is $0$ if $|x/c|\ge 1$.  If $-1<x/c<1$, then 
$$\begin{align}
\int_{-1}^1 \frac{k}{\sqrt{1-k^2}}\,\text{sgn}(k+x/c)\,dk&=-\int_{-1}^{-x/c} \frac{k}{\sqrt{1-k^2}}\,dk+\int_{-x/c}^1 \frac{k}{\sqrt{1-k^2}}\,dk\\\\
&=2\sqrt{1-(x/c)^2}\tag 5
\end{align}$$
Finally, using $(5)$ in $(4)$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{F(\omega)=\begin{cases}\sqrt{c^2-x^2}&,|x|<c\\\\0&,|x|\ge 0\end{cases}}$$
