Explanation about the transformation

Question

Is there any explanation for : What is the transformation used? $$ \frac{\mathit{\lambda}}{\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}}\mathop{\int}\limits_{t0}\limits^{t}{{\mathrm{(}}{t}\mathrm{{-}}{s}{\mathrm{)}}^{{q}\mathrm{{-}}{1}}}{x}_{0}\mathrm{(}s\mathrm{)}ds $$ How did become like this $$ \frac{{x}_{0}\hspace{0.33em}\mathit{\lambda}}{\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}}\mathop{\int}\limits_{0}\limits^{1}{{\mathrm{(}}{t}\mathrm{{-}}{t}_{0}}{\mathrm{)}}^{{2}{q}\mathrm{{-}}{1}}{\mathrm{(}}{1}\mathrm{{-}}\mathit{\sigma}{\mathrm{)}}^{{q}\mathrm{{-}}{1}}{\mathit{\sigma}}^{{q}\mathrm{{-}}{1}}{d}\mathit{\sigma} $$ Where: $$ {x}_{0}{\mathrm{(}}{t}{\mathrm{)}}\mathrm{{=}}\frac{{x}_{0}{\mathrm{(}}{t}\mathrm{{-}}{t}_{0}{\mathrm{)}}^{{q}\mathrm{{-}}{1}}}{\mathit{\Gamma}{\mathrm{(}}{q}{\mathrm{)}}} $$

Answer 1

Just put $x_0(t)$ inside the integral and see where it goes $$\begin{align}\frac{\lambda}{\Gamma(q)}\int_{t_0}^t(t-s)^{q-1} x_0(s)ds&=\frac{\lambda}{\Gamma(q)}\int_{t_0}^t(t-s)^{q-1}\frac{(s-t_0)^{q-1}}{\Gamma(q)}x_0\,ds\\ &=\frac{x_0\;\lambda}{\Gamma(q)\Gamma(q)}\int_{t_0}^t\left[(t-s)(s-t_0)\right]^{q-1}ds\end{align}$$ Now let $s:=\sigma(t-t_0)+t_0$, then $$\begin{align}t-s&=(1-\sigma)(t-t_0)\\ s-t_0&=\sigma(t-t_0)\\ ds&=(t-t_0)d\sigma\end{align}$$ and the bounds of integration changes to $(0,1)$: $$\int_{t_0}^t\left[(t-s)(s-t_0)\right]^{q-1}ds=\int_0^1\left[\sigma(1-\sigma)\right]^{q-1}(t-t_0)^{2(q-1)}(t-t_0)d\sigma\\=\int_0^1\sigma^{q-1}(1-\sigma)^{q-1}(t-t_0)^{2q-1}d\sigma$$