= Abel integral equation of fractional order
{c}
{title2=$\int_0^t\Phi(\tau)(t-\tau)^{-\alpha}d\tau=f(t),\quad0<\alpha<1$}
For an absolutely continuous $f$ with $f(0)=0$, this causal <Abel integral equation> has inversion
$$
\Phi(t)=\frac1{\Gamma(\alpha)\Gamma(1-\alpha)}\int_0^tf'(\tau)(t-\tau)^{\alpha-1}d\tau.
$$
Under Laplace-transform growth hypotheses, the <convolution theorem for Laplace transforms> converts the original kernel into $\Gamma(1-\alpha)s^{\alpha-1}$. Since $\mathcal L[f']=s\mathcal L[f]$, the transformed inverse is $s^{-\alpha}\mathcal L[f']/\Gamma(1-\alpha)$, 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 $\int_0^tf'=f(t)$. The same convolution identity proves uniqueness in that class. For example $f(t)=t$ gives $\Phi=t^\alpha/[\Gamma(\alpha+1)\Gamma(1-\alpha)]$.
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