Abel integral equation ($g(x)={\int }_x^{b}f(y)(y-x{)}^{-1/2}dy$)

This square-root-kernel integral equation is inverted by integrating $g(x)/\sqrt{x-z}$ over $z<x<b$ and interchanging integrations. The inner integral over $z<x<y$ equals $\pi$, so ${\int }_z^{b}g(x)/\sqrt{x-z}dx=\pi {\int }_z^{b}f(y)dy$. Differentiating recovers $f(z)=-{\pi }^{-1}d[{\int }_z^{b}g(x)/\sqrt{x-z}dx]/dz$. The changes of variables $x=p^2$, $y={\eta }^2$ put the angular-ray equation into this form.

Suppose $h$ is continuously differentiable on $[0,L]$. For $0<x\le L$, the improper integral $g(x)={\int }_0^{x}h(t)(x-t{)}^{-1/2}dt$ exists, but substituting directly into the ordinary Leibniz integral rule produces an undefined endpoint term. First use integration by parts to obtainThe new endpoint term at $t=x$ vanishes, and the differentiated kernel is locally integrable. Differentiating gives the displayed formula. One rigorous justification rewrites the second term as ${\int }_0^x[{\int }_0^s{h}^{\prime }(t)(s-t{)}^{-1/2}dt]ds$ by Fubini's theorem; the inner integral is continuous for $s>0$, so the fundamental theorem of calculus applies. Moreover $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$: if $h(0)\ne 0$, the right difference quotient at zero diverges rather than defining a finite derivative.

For an absolutely continuous $f$ with $f(0)=0$, this causal Abel integral equation has inversionUnder Laplace-transform growth hypotheses, the convolution theorem for Laplace transforms converts the original kernel into $\Gamma (1-\alpha )s^{\alpha -1}$. Since $\mathcal{L}[f^{\prime }]=s\mathcal{L}[f]$, the transformed inverse is $s^{-\alpha }\mathcal{L}[f^{\prime }]/\Gamma (1-\alpha )$, giving the formula. It also holds locally in the corresponding integrable convolution class: convolving the two power kernels produces the constant $B(\alpha ,1-\alpha )=\Gamma (\alpha )\Gamma (1-\alpha )$ by the beta-gamma identity, so substitution recovers ${\int }_0^t{f}^{\prime }=f(t)$. The same convolution identity proves uniqueness in that class. For example $f(t)=t$ gives $\Phi =t^{\alpha }/[\Gamma (\alpha +1)\Gamma (1-\alpha )]$.