Abel integral equation of fractional order (${\int }_0^t\Phi (\tau )(t-\tau {)}^{-\alpha }d\tau =f(t),0<\alpha <1$)

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 $\Phi =t^{\alpha }/[\Gamma (\alpha +1)\Gamma (1-\alpha )]$.