# Inverse fourier transform of $f(x)=\frac{\text{u}(x)}{\sqrt{1-x^2}}$

How to calculate (or derive) the inverse Fourier transform of

$$f(x)=\frac{\text{u}(x)}{\sqrt{1-x^2}}$$

where $u(x)$ is the rectangular function? I know that $f(x)$ is the Fourier transform of the Bessel function, so its inverse must be the Bessel function, but how to show?

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Well, have you tried computing $g(k) = \tfrac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {f(x){e^{ixk}}dx}$? – glebovg Oct 31 '12 at 23:21

I assume "the rectangular function" is the indicator function of the interval $[-1,1]$. Thus you want to calculate $$F(s) = \int_{-1}^1 \frac{e^{isx}}{\sqrt{1-x^2}} \ dx$$ Expand it in a power series: $e^{isx} = \sum_{n=0}^\infty (isx)^n/n!$ (converging uniformly on this interval for any fixed $s$). Since $1/\sqrt{1-x^2}$ is integrable, we can interchange integral and sum. Now let $$L_n = \int_{-1}^{1} \frac{x^n \ dx}{\sqrt{1-x^2}}$$
By symmetry, $L_n = 0$ for odd $n$. So now assume $n = 2m$ is even. The substitution $x = \cos(t)$ gives us $$L_{2m} = \int_0^{\pi} \cos(t)^{2m}\ dt = \frac{1}{2} \int_0^{2\pi} \left(\dfrac{e^{it} + e^{-it}}{2}\right)^{2m}\ dt = 2^{-2m-1} \sum_{j=0}^{2m} {2m \choose j} \int_0^{2\pi} e^{2(j-m)it} \ dt$$ All the integrals on the right side are $0$ except for the case $j=m$. Thus we get $$L_{2m} = 2^{-2m} \pi {2m \choose m} = 2^{-2m} \pi \dfrac{(2m)!}{(m!)^2}$$ We conclude that $$F(s) = \sum_{m=0}^\infty 2^{-2m} \pi \frac{(-s^2)^{m}}{(m!)^2} = \pi J_0(s)$$