# Hankel transform. Change of variables.

I stuck with a rather simple question, but i can't fix it. Hope you can help me.

Hankel transform for a cylindrically symmetric problem is given by the following formula $g(\rho) = 2\pi \int \limits_{0}^{\infty} r f(r) J_l(2 \pi r \rho) dr$

Inverse transform looks the same. It is obvious that in a real system we deal with a bounded region $[0,r_{max}]$. Reading an article concerning numerical implementation of HT I met the change of variables: $x=r/b$ and $y=\rho/\beta$

where $b$ and $\beta$ are maximum values of spatial ($r$) and frequency ($\rho$) domain, respectively.

So, authors obtained the following expression

$g(y) = 2\pi \gamma \frac{b}{\beta} \int \limits_0^1 x f(x) J_l (2 \pi \gamma x y)dx$

where $\gamma = b \beta$

It is not obvious for me how to move from $g(\rho) = g(\beta y)$ to $g(y)$. And the same with $f(bx) \to f(x)$ under the integal.

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I found an answer. This change of variables is described in Goudman's Introduction to Fourier Optics. – jacksonslsmg4 Sep 28 '12 at 8:30

SO, $r=bx$, where $b$ is some constant, then $dr=b\cdot dx$ (meaning exactly that the differentiate of $r$ by $x$, $\displaystyle\frac{dr}{dx} = b$). So, $$g(y)=2\pi \underset{bx=0}{\overset{bx=b}\int} bx f(bx)J_l(2\pi bx y)\cdot b dx$$ Moving $g(\rho)$ to $g(y)$ was just substituting $y$ in the place of the variable. But, for moving $f(bx)$ to $f(x)$, I guess, needs some assumption on $f$ (for example, being linear), else it's not changing..
No, it's not linear. In general $f(r)$ describes electric field distribution in a laser pulse. i don't understand how they move from $r$ to $x$ in $f$'s argument. – jacksonslsmg4 Sep 26 '12 at 11:36