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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.

Vivek Kaushik
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Kirin171
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    Doing the inverse transform will just get you back to the original function (which is indeed a (semi-)circle). It seems like the question is asking you to set up the integral for the inverse transform, not actually calculate it. – Carmeister May 10 '17 at 03:18
  • This makes sense, thanks – Kirin171 May 10 '17 at 03:22

1 Answers1

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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

$$\bbox[5px,border:2px solid #C0A000]{F(\omega)=\begin{cases}\sqrt{c^2-x^2}&,|x|<c\\\\0&,|x|\ge 0\end{cases}}$$

Mark Viola
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