Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need an inversion formula for the Abel transform $$ F(y) = 2\int_y^\infty\frac{f(r)r\,dr}{\sqrt{r^2-y^2}}. $$ Hint: The inversion formula found on Wikipedia appears to be incorrect. The "Verification of the inverse Abel transform" section makes the mistake to formally differentiate the integrand, but neglecting that the lower bound of the integral depends on $y$ as well and need to be considered in the derivative.

Thanks in advance.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

If we write the Abel transform as operator $\mathcal A$ (and consider the input and output functions as even) and the Fourier-Transform as $\mathcal F$, then $$ \mathcal A^{-1} = \mathcal{F}^{-1}\mathcal{A}\ \mathcal{F}\,. $$ Here's the proof: We define the operators $\mathcal M$ and $\mathcal{I}$ by \begin{align} \mathcal{M}f(x,y) &:= f\left(\sqrt{x^2+y^2}\right), \\ \mathcal{I}g(y) &:= \int_{-\infty}^\infty g(x,y)dx \end{align} Combining these operators yields $$ \mathcal{I}\mathcal{M}f(y) = 2\int_0^\infty f\left(\sqrt{x^2+y^2}\right) dx = 2\int_{|y|}^\infty \frac{f(r)r dr}{\sqrt{r^2-y^2}} = \mathcal{A}f(y) $$ and hence $\mathcal{I}\mathcal{M} = \mathcal{A}$. Applying the Fourier transform to $\mathcal{I}g$ gives us $$ \mathcal{F}\mathcal{I}g(\eta) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} g(x,y)e^{-2\pi iy\eta} dx dy = (\mathcal{F}g)(0,\eta), $$ which is also known as the Fourier slice theorem. If $g$ is a circular function (i. e. $g(x,y) = g(\sqrt{x^2+y^2},0)$), then it's Fourier transform is also circular. Therefore, in this case we have $$ \mathcal{F}g(\xi,\eta) = \mathcal{F}\mathcal{I}g\left(\sqrt{\xi^2+\eta^2}\right) = \mathcal{M}\mathcal{F}\mathcal{I}g(\xi,\eta). $$ Hence for circular $g$ we get $$ (\mathcal{M}\mathcal{F}\mathcal{I})^2g =\mathcal{F}\mathcal{F}g =g, $$ if we use the isometric variant of the Fourier transform (otherwise there would simply be another constant factor). Now we plug in $g:=\mathcal{M}f$ and get $$ \mathcal{M}f = \mathcal{M}\mathcal{F}\underbrace{\mathcal{I}\mathcal{M}}_{=\mathcal A}\mathcal{F}\underbrace{\mathcal{I}\mathcal{M}}_{=\mathcal A}f = \mathcal{M}\mathcal{F}\!\mathcal{A}\mathcal{F}\!\mathcal{A}f. $$ Since $\mathcal{M}$ is injective we can take away the $\mathcal{M}$ on the left side of the expressions and get $ f = \mathcal{F}\!\mathcal{A}\mathcal{F}\!\mathcal{A}f $. Since $\mathcal{A}\mathcal{F}\!\mathcal{A}f$ is even, the $\mathcal{F}$ left of it can be replaced by $\mathcal{F}^{-1}$ and hence we get $ f = \mathcal{F}^{-1}\!\mathcal{A}\mathcal{F}\!\mathcal{A}f $ which holds true, even if the coefficient in the definition of the Fourier transform is chosen to be different from $1$. This completes the proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.