 Solution

Treat the left side as a causal convolution with $t^{-\alpha }$. Under the usual Laplace-transform hypotheses, the convolution theorem for Laplace transforms givesFor an absolutely continuous $f$ with $f(0)=0$, the derivative transform is $\mathcal{L}[f^{\prime }]=s\hat{f}$. ConsequentlySince $\mathcal{L}[t^{\alpha -1}]=\Gamma (\alpha )s^{-\alpha }$, inverting the product provesHere $0<\alpha <1$ makes both power kernels locally integrable. For example assume exponential growth bounds sufficient for the transforms and a regularity class where the convolution is defined. This is the Abel integral equation of fractional order; the initial value removes the extra boundary term in the derivative transform.

For $f(t)=t$, the remaining integral is ${\int }_0^t(t-\tau {)}^{\alpha -1}d\tau =t^{\alpha }/\alpha$. ThusSubstitution into the original equation checks the result: scaling $\tau =tu$ gives $tB(\alpha +1,1-\alpha )/[\Gamma (\alpha +1)\Gamma (1-\alpha )]=t$ by the beta-gamma identity.