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I am trying to show that the Hankel transform of the rect function is the jinc function.

The Hankel transform is defined as follows: $$ F(q) = \mathbb{H}\left\{f(r)\right\} = 2\pi\int_0^\infty f(r) \hspace{1 mm} J_0\left(2\pi rq\right) \hspace{1 mm} r \hspace{1 mm} dr $$ where $J_0$ is the bessel function of the first kind zeroth order.

$$ \text{rect}(x) = \left\{ \begin{array}{ll} 1 & |x| \leq 1/2 \\ 0 & |x| \gt 1/2 \\ \end{array} \right. $$

jinc is defined as $$ \text{jinc}(q) = \frac{J_1(\pi q)}{2q} $$ where $J_1$ is the bessel function of the first kind, first order.

Thank you all very much for your help.

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I think you would rather define $f(r)$ as follows:

$$ f(r) = \begin{cases} 1 & r \in [0,1] \\ 0 & r>1 \\ \end{cases}$$

Then use the fact that

$$ \int dx \: x J_0(x) = - x J_0'(x) + C = x J_1(x) + C $$

where $C$ is a constant of integration. The integration follows directly from the differential equation for $J_0$:

$$x J_0'' + J_0' + x J_0 = (x J_0')' + x J_0 = 0 $$

Then your Hankel transform becomes

$$ 2 \pi \int_0^1 dr \: r \, J_0(2 \pi q r) = \frac{2 \pi}{(2 \pi q )^2} 2 \pi q \, J_1(2 \pi q) = \frac{J_1(2 \pi q)}{q} $$

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  • $\begingroup$ Oh my gosh, you're right. Sorry; I didn't even get the problem right. $\endgroup$ – NicNic8 Jan 22 '13 at 5:29

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