# Inverse transform of a modified Abel transform

I have been struggling for 6 months on finding the analytical inverse transform of a transformation below:

$$F(y,k) = 2 \int_y^{\infty}\cos\left(ka\sqrt{r^2-y^2}\right) f(r,k) \frac{r}{\sqrt{r^2-y^2}}\ \mathrm{d}r$$

where $a$ is a known constant and $\lim_{r\to\infty}f(r,k) = f'(r,k) = 0$ and all functions are continuous.

It has a similar form of Abel transform, except that it has the cosine factor in the integrand. Any help will be appreciated!

• Joonas Ilmavirta consider a different generalisation of Abel transform in the paper users.jyu.fi/~jojapeil/pub/disk.pdf, section 3, especially lemma 10. He has Chebyshev polynomials instead of cosine term, but the references or techniques might be useful. – Tommi Feb 18 '15 at 12:33
• I have read the reference and it refers to another paper in scitation.aip.org/content/aip/journal/jap/34/9/10.1063/… for the inverse transform derivation. In the paper (the AIP one), he multiplied the left hand side with some factors, integrate it, and everything's going well. The problem is that he didn't explain explicitly how to get the factor. I also tried the similar factor in this case, but it didn't work. – Firman Feb 19 '15 at 16:31

Note that $F$ also depends on $a$:

\begin{align}F(y,k,a) &= 2 \int_0^{\infty} du \, \cos{(k a u)}\, f \left (\sqrt{u^2+y^2},k \right ) \\ &= \int_{-\infty}^{\infty} du \, e^{i k a u} \, f \left (\sqrt{u^2+y^2},k \right )\\ &= \frac{1}{k} \int_{-\infty}^{\infty} dv \, e^{i a v} \, f \left (\sqrt{k^2 v^2+y^2},k \right )\end{align}

because $f$ with the argument in the above integrand is even. Thus,we have a Fourier transform relationship:

$$f \left (\sqrt{k^2 v^2+y^2},k \right ) = \frac{k}{2 \pi} \int_{-\infty}^{\infty} da \, e^{-i a v} F(y,k,a)$$

or

$$f(r,k) = \frac{k}{2 \pi} \int_{-\infty}^{\infty} da \, e^{-i (a/k) \sqrt{r^2-y^2}} F(y,k,a)$$

• Thank you for the answer. I forgot to mention in the question that $a$ is a known constant. – Firman Feb 19 '15 at 12:36