I was dealing with the integral operator
$$\tilde{I}^{\alpha}_{-, 2} f \left( r \right) = \int\limits_{r}^{\infty} \frac{f (s, r)}{(s^2 - r^2)^{1 - \alpha}} s \mathrm{d}s,$$
for "nice" functions $f: (0, \infty) \times (0, \infty) \rightarrow \mathbb{R}$. This operator looks incredibly like the modified Erdelyi-Kober fractional derivative of order $\alpha$. I was wondering if one could get back, say $f(r,r)$ from the knowledge of $\tilde{I}^{\alpha}_{-, 2} f (r)$?
One idea that I had was to look at the operator $I^{\alpha, 1}_{-, 2}f (t, r) = \int\limits_{t}^{\infty} \frac{f(s, r)}{(s^2 - t^2)^{1 - \alpha}} s \mathrm{d}s$, and see that $\lim\limits_{t \rightarrow r} I^{\alpha, 1}_{-, 2} f \left( t, r \right) = \tilde{I}^{\alpha}_{-, 2} f(r)$. Then use the fractional derivatives to get back $f(s, r)$ first from the left-hand-side and then put $s = r$. However, at this point it is a very crude approach, since I do not feel that the limit and the fractional derivative can be interchanged as such!
Any insights into this are well appreciated!
PS: Is there any way to check the injectivity of the operator? If it is not injective, then it does not make sense to ask for a left-inverse of it!