I have been battling a variant of Abel's integral for a few days. I found a similar question here, in the sense that this question asks about the Beta function, but the integral is different. Interestingly, I found exactly my same integral at this other question, but there are no answers yet. My problem arises in the derivation of the inverse Radon transform (needed for the reconstruction algorithm for CT scans and many other fields). Radon himself (1917) makes a reference to Abel and sets up the Abel-type integral in this form $$ \bar{F}(q) = \int_u^\infty {\bar f(v) \over \sqrt{v-u}} dv. $$ Using more neutral variable names, this is almost the same as Abel's integral in this form: $$ f(s) = \int_0^s {g(t) \over (s-t)^\alpha} dt, $$ where for my case $\alpha = 1/2$.
Kanwal (Linear Integral Equations, 1997) provides an excellent step-by step explanation of how this integral is solved by transforming it into a Beta function integral. His solution makes perfect sense. Quoting Sneddon (1966) he also says that a similar method and solution apply to the variant $$ f(s) = \int_s^b {g(t) \over \sqrt{t-s}} dt, $$ where $b$ can be $\infty$. The actual final solution for $g(t)$ is provided by Radon, Kanwal, Sneddon, and also by Herman (2009), a more applied reference specific to CT scans theory and practice. But my question is how the variant can be expressed as a Beta function. In a sense, it is a trivial question about variable substitution and integral limits, but there is a bit of subtlety that my non-mathematician mind is finding challenging. I will outline Kanwal's solution to Abel's integral, and will then follow similar steps to end up with a Beta integral whose limits are not standard. The question, therefore, will end up being whether I made an error in the variable substitution or perhaps whether the odd limits are not actually a problem.
Multiply by $ds/\sqrt{u-s}$ the second equation above and integrate on both sides (assuming $\alpha = 1/2$): $$ \int_0^u {f(s)\over\sqrt{u-s}} ds = \int_0^u \left[ \int_0^s {g(t) dt\over\sqrt{s-t}} \right] {ds\over\sqrt{u-s}}. $$ The integrals on the RHS, together, represent the integration over a triangular 2D region, as shown on the left in the figure below. Inverting their order gives the situation on the right. The inner integral is visualised with 3 thin arrows, whereas the outer integral with the thick arrow. Colours match the directions. The limits of the now inverted integrals are slightly different: $$\int_0^u {f(s)\over\sqrt{u-s}} ds = \int_0^u \left[\int_t^u {ds\over\sqrt{u-s}\sqrt{s-t}} \right] g(t) dt. $$ In particular, the new limits of the inner integral are really important for obtaining the Beta function.
If we now focus on the inner integral, Kanwal introduces the transformation $$ y = {u-s\over u-t}, \qquad\text{such that}\qquad dy = -{ds\over u-t}. $$ The limits transform as $$ s = t \quad \Rightarrow \quad y = 1, \qquad\qquad s = u \quad \Rightarrow \quad y = 0. $$ Defining also $(1-y)$ and substituting everything into the inner integral, we get: $$-\int_1^0 {dy\over y^{1/2} (1-y)^{1/2}} = \int_0^1 {dy\over y^{1/2} (1-y)^{1/2}} = {\pi \over \sin{\pi/2}} = \pi. $$ From there the derivation continues and after integration by parts eventually obtains an expression for $g(t)$ in terms of $f(0)$ and the integral of the derivative of $f(s)$ between $0$ and $t$.
If we now look at the variant, we have $$\int_0^u {f(s)\over\sqrt{s-u}} ds = \int_0^u \left[ \int_s^b {g(t) dt\over\sqrt{t-s}} \right] {ds\over\sqrt{s-u}}$$ $$\int_0^u {f(s)\over\sqrt{s-u}} ds = \int_0^b \left[\int_0^t {ds\over\sqrt{s-u}\sqrt{t-s}} \right] g(t) dt, $$ corresponding to the figure:
I am pretty sure this second figure is not right but I don't know how to fix it: should $u = b$? Can both be $\infty$?
At this point I tried defining $y$ as above and ended up with $$ -\int_0^{u\over u-t} {dy\over y^{1/2} (1-y)^{1/2}}. $$ If $u \rightarrow \infty$ perhaps this is not wrong since the upper limit would become 1, but I am not sure if I arrived at the integral itself correctly.
My next step would be to start delving into the Beta function and to think through the second derivation in order to understand what's going on a bit better instead of replicating blindly the steps of the first derivation, but I am a bit pressed for time and thought that maybe someone can see how to fix this easily. Many thanks in advance.
Edit
I am starting to wonder if I should have posted this question in the physics stack exchange. Sorry for cluttering, I'll think better next time. However, now that I am here I might as well continue. I made an error with the integration limits in the last step. The correct integral is $$ \int_{u\over u-t}^1 {dy\over y^{1/2} (1-y)^{1/2}}, $$ where I also inverted the direction. Rather than blindly letting $u \rightarrow \infty$, the original problem had $b \rightarrow \infty$, which corresponds to the direction of $t$. Thus, if I let $t \rightarrow \infty$ the lower limit will become $0$! I do not know whether this is legitimate, but the implication would appear to be that the first and second types of the Abel integral give exactly the same result: when $\alpha = 1/2$ the result is $\pi$ in both cases. Is that correct??
Hopefully this is a slightly more fundamental type of question that might stimulate a response :)
Edit 2
Actually, I just looked up again Kanwal and Sneddon, and the answer for the second type is supposed to be $-\pi$ rather than $\pi$. This is reassuring but it's also telling me that I am still doing something wrong.
The second figure above is the source of my problem. I am going to edit this answer without saving the prior version, hopefully that's OK. The correct figure, I believe, is:
This leads to the following limits:
$$\int_u^b {f(s)\over \sqrt{s-u}}ds = \int_u^b \left[ \int_s^b {g(t)dt \over \sqrt{t-s}} \right] {ds\over \sqrt{s-u}} $$
$$\qquad\qquad\qquad\quad = \int_u^b \left[ \int_u^t {ds \over \sqrt{t-s}\sqrt{s-u}} \right] g(t)dt. $$
Focusing on the inner integral, let $$ y = {u-s\over u-t}\quad \Rightarrow \quad dy = -{ds\over u-t}\quad \Rightarrow \quad (s-u )=-y(u-t)$$ $$ 1-y = - {t-s\over u-t} \quad \Rightarrow \quad (t-s) = -(1-y)(u-t) $$ $$ s = u \quad \Rightarrow \quad y = 0, \qquad\qquad s = t \quad \Rightarrow \quad y = 1. $$
Substituting, $$ \int_u^t {ds \over \sqrt{t-s}\sqrt{s-u}} = -\int_0^1 {dy \over y^{1/2} (1-y)^{1/2}} = -{\pi\over \sin {\pi\over2}} = -\pi, $$ I don't have a geometrical sense for what the Beta function is or does but I guess that's for another day.
I was able to finally figure out the domain after I found the original reference for these integrals: Srivastav, R P (1963). A note on certain integral equations of Abel-type. Proceedings of the Edinburgh Mathematical Society, 13(3):271–272. Link