Abel's integrate equation (Laplace transform)

Question

I am stuck in my textbook; it states: Given Abel's integral equation: $$f(t)=\int_{0}^{t} \frac {\phi(\tau)}{(t-\tau)^{\alpha}} d\tau$$ It is clear that the Laplace transform can be represented as follows: $$\bar{\phi }(s) = \frac{1}{\Gamma(1-\alpha)}s^{1-\alpha}\bar{f}(s)$$ Perhaps for most readers it is clear, but I can't get to it. Perhaps I am confused by the number of $\phi's$ and $\alpha's$. Perhaps somebody is able to put me on the right track?

Answer 1

After 4 hours of thinking and trying, I think I found the solution. I hope it is correct, if not, please don't hesitate to correct it. The RHS of the equation is the convolution of $\phi (t)$ and $\frac{1}{t^{-\alpha}}$ and so$$F(s) = \mathcal{L}[f(t)] = \mathcal{L}[\phi(t)]\mathcal{L}\bigg[\frac{1}{t^\alpha}\bigg] =\bar{\phi }(s)\int_{0}^{\infty}\frac {e^{-st}}{t^{\alpha}}dt$$By taking $s=x>0$ and putting $\tau = xt$ to obtain$$\int_{0}^{\infty}\frac {e^{-xt}}{t^{\alpha}} = x^{-(1-\alpha)} \int_{0}^{\infty} \tau^{(1-\alpha)-1}e^{-\tau}d\tau$$ $$= x^{-(1-\alpha)}\Gamma (1-\alpha)$$

If we continue this we note that: $$\mathcal{L}\bigg[\frac{1}{t^\alpha}\bigg] = s^{-(1-\alpha)}\Gamma (1-\alpha)$$ Hence: $$\bar{\phi }(s) = \frac{1}{\Gamma(1-\alpha)}s^{1-\alpha}\bar{f}(s)$$