I have been reading this solution. In part of the solution there is an integral which leads to beta function but I don't know how. Here it is
\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} Why is second integral equal to the third one and how is it equal to beta function?!
$\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}dx=\text{Beta}(\alpha, \beta)$ if $\alpha,\beta>0$. For that little problem, let $u=\tau-x$, so that $du=d\tau$:
$$\begin{split}\int_x^t \frac{1}{(t-\tau)^{1-\alpha}(\tau-x)^\alpha}d\tau &=\int_0^{t-x}\frac 1{(t-u-x)^{1-\alpha}u^\alpha}du\end{split}$$
Assume $t-x=1$, then
$$\begin{split}\int_0^{1}\frac 1{(1-u)^{1-\alpha}u^\alpha}du&=\int_0^1 u^{-\alpha+1-1}(1-u)^{\alpha-1}du\\ &=\text{Beta}(1-\alpha, \alpha)\end{split}$$