Use of a Laplace Transform to solve Abel's Integral Equation I am following an exercise where I have to solve Abel's integral equation:
$$f(t)=\int_{0}^{t} \frac {\phi(\tau)}{(t-\tau)^{\alpha}} d\tau$$
I have taken the first step and shown that:
$$\bar{\phi }(s) = \frac{1}{\Gamma(1-\alpha)}s^{1-\alpha}\bar{f}(s)$$
However, I now have to assume a function $\psi$ such that $\phi=d\psi/dt$ and $\psi(0)=0$.  Using my above result I have find the Laplace transform of $\psi(t)$ and then show that 
$\phi(t)=\frac{\sin \pi \alpha}{\pi}\frac{d}{dt}\int^t_0\frac{f(\tau)}{(t-\tau)^{1-\alpha}}d\tau$
Could anyone provide some pointers on how I would do this?  I am guessing that the sine term will appear due to the identity $1/\Gamma(z)\Gamma(1-z)=\frac{\sin \pi z}{\pi}$ but not sure how exactly to carry it all out.
 A: We assume that $0<\alpha <1$ and $f(0)=0$ for Abel's integral equation$$
f(t)=\int_{0}^{t} \frac {\phi(\tau)}{(t-\tau)^{\alpha}} d\tau.
$$
We obtain$$
\phi(t)=\frac{\sin \pi \alpha}{\pi}\frac{d}{dt}\int^t_0\frac{f(\tau)}{(t-\tau)^{1-\alpha}}d\tau.\tag{1}
$$
Then integrating by parts we have
\begin{align}
\int^t_0\frac{f(\tau)}{(t-\tau)^{1-\alpha}}d\tau&=\left.-\frac{1}{\alpha }(t-\tau)^\alpha f(\tau)\right|_0^t+\frac{1}{\alpha }\int_0^t (t-\tau)^\alpha f^\prime(\tau)d\tau\\
&=\frac{1}{\alpha }\int_0^t (t-\tau)^\alpha f^\prime(\tau)d\tau.
\end{align}
Now by differentiation under the integral sign we have 
\begin{align}
\phi(t)&=\frac{\sin \pi \alpha}{\pi}\frac{d}{dt}\left(
\frac{1}{\alpha }\int_0^t (t-\tau)^\alpha f^\prime(\tau)d\tau\right)\\
&=\frac{\sin \pi \alpha}{\pi}\frac{1}{\alpha }\left((t-t)^\alpha f^\prime(t) +\int_0^t \alpha (t-\tau)^{\alpha -1}f^\prime(\tau)d\tau \right)\\
&=\frac{\sin \pi \alpha}{\pi}\int_0^t (t-\tau)^{\alpha -1}f^\prime(\tau)d\tau.
\end{align}
Thus we have the solution
$$
\phi(t)=\frac{\sin \pi \alpha}{\pi}\int_0^t (t-\tau)^{\alpha -1}f^\prime(\tau)d\tau
$$
for Abel's integral equation
$$
f(t)=\int_{0}^{t} \frac {\phi(\tau)}{(t-\tau)^{\alpha}} d\tau.
$$
EDIT: 
Derivation of $(1)$: 
Start with $$
f(\tau)=\int_0^\tau \frac{\phi(x)}{(\tau-x)^\alpha}dx.\tag{2}$$
Multiply both sides of $(2)$ by $\frac{1}{(t-\tau)^{1-\alpha }}$ and integrate in $\tau$ from $0$ to $t$, then we have
\begin{align}
\int_0^t\frac{f(\tau)}{(t-\tau)^{1-\alpha }}d\tau&=\int_0^t \frac{1}{(t-\tau)^{1-\alpha }}\left(\int_0^\tau\frac{\phi(x)}{(\tau-x)^\alpha }dx \right)d\tau\\
&=\int_0^t\phi(x)\left(\int_x^t \frac{1}{(t-\tau)^{1-\alpha }(\tau-x)^\alpha }  \right)dx\\
&=B(\alpha ,1-\alpha )\int_0^t \phi(x)dx.
\end{align}
Therefore we have
$$
\int_0^t \phi(x)dx=\frac{\sin \pi \alpha }{\pi}\int_0^t\frac{f(\tau)}{(t-\tau)^{1-\alpha }}d\tau
$$
and $(1)$.  
Solution by use of Laplace transform: 
Considering Laplace transform of both sides of given integral equation we have
$$
\int_0^\infty e^{-st}\left(\int_{0}^{t} \frac {\phi(\tau)}{(t-\tau)^{\alpha}} d\tau\right)dt=\int_0^\infty e^{-st}f(t)dt.$$
\begin{align}
\text{LHS}&=\int_0^\infty \phi(\tau) d\tau\int_\tau^\infty \frac{e^{-st}}{(t-\tau)^\alpha }dt=\int_0^\infty \phi(\tau) d\tau\int_0^\infty \frac{e^{-s(u+\tau)}}{u^\alpha }du\\
&=\int_0^\infty e^{-s\tau}\phi(\tau) d\tau\int_0^\infty \frac{e^{-su}}{u^\alpha }du\\
&=\int_0^\infty e^{-s\tau}\phi(\tau) d\tau\cdot \frac{\Gamma(1-\alpha)}{s^{1-\alpha}}.
\end{align}
Thus we have \begin{align}
\int_0^\infty e^{-s\tau}\phi(\tau) d\tau&=\frac{s^{1-\alpha}}{\Gamma(1-\alpha)}\int_0^\infty e^{-st}f(t)dt.
\end{align}
Integration by parts gives
$$
\int_0^\infty e^{-st}f(t)dt=\frac{1}{s}\int_0^\infty e^{-st}f^\prime(t)dt.
$$
So we have \begin{align}
\int_0^\infty e^{-s\tau}\phi(\tau) d\tau&=\frac{1}{\Gamma(1-\alpha)}\cdot \frac{1}{s^\alpha }\int_0^\infty e^{-st}f^\prime(t)dt\\
&=\frac{1}{\Gamma(1-\alpha)\Gamma(\alpha)}\left(\int_0^\infty e^{-st}t^{\alpha -1}dt\right)\left(\int_0^\infty e^{-st}f^\prime(t)dt\right)\\
&=\frac{\sin \pi\alpha}{\pi} \mathcal{L}\{s; t^{\alpha -1}\}\mathcal{L}\{s; f^\prime(t)\},
\end{align}
where $\mathcal{L}$ denotes Laplace transform.
By the well known formula$$
\mathcal{L}\{f\}\mathcal{L}\{g\}=\mathcal{L}\{f*g\},$$
where $(f*g)(t)=\int_0^t f(t-\tau)g(\tau)d\tau$, we have
\begin{align}
\mathcal{L}\{\phi\}&=\frac{\sin \pi\alpha}{\pi}\mathcal{L}\{t^{\alpha -1}*f^\prime \}=\frac{\sin \pi\alpha}{\pi}\mathcal{L}\left\{\int_0^t \frac{f^\prime(\tau)}{(t-\tau)^{1-\alpha }}d\tau\right\}
\end{align}
and we have $$
\phi(t)=\frac{\sin \pi\alpha}{\pi}\int_0^t \frac{f^\prime(\tau)}{(t-\tau)^{1-\alpha }}d\tau
$$
by inversion formula.
