Multivariable integral over a simplex

Let $p$ be a positive integer, let $B > A >0$ and let $\beta >0 $ and $\beta \neq 1$. With a help of Mathematica (ie using elementary integration and consecutive simplifications) I have shown that :
\begin{eqnarray}
&&I^{(A,B)}_p :=
\int\limits_{A\le \xi_0 \le \cdots \le \xi_{p-1} \le B} \prod\limits_{j=0}^p \left(\xi_{j-1} – \xi_j\right) \cdot \prod\limits_{j=0}^{p-1} \frac{d \xi_j}{\xi_j^{2 \beta}} = (-1)^p \frac{\frac{\pi}{2 \beta-2}}{(2\beta-2)^{2 p} (p!)^2 \sin(\frac{\pi}{2 \beta-2})} \cdot \\
&&
\sum\limits_{q=0}^p \binom{p}{q} A^{(2-2\beta)(p-q)} B^{(2-2\beta) q} \left(A \binom{p}{q+\frac{1}{2\beta-2}} – B\binom{p}{q-\frac{1}{2\beta-2}}\right)
\end{eqnarray}
subject to $\xi_{-1}=A$ and $\xi_{p}=B$.
Is there a way to show this result using some different method?

Note, that setting $\beta =0$ we readily obtain
\begin{equation}
I_{p}^{(A,B)}\left(0\right) := (-1)^p\frac{(A-B)^{2p+1}}{(2p+1)!}
\end{equation}
as it should be.
Likewise setting $\beta=2$ we get:
\begin{equation}
I_{p}^{(A,B)}\left(2\right) := (-1)^p\frac{(A-B)^{2p+1}}{(2p+1)! (A B)^{2 p}}
\end{equation}

Below we also give a similar integral to the one above.
\begin{eqnarray}
&&J^{(A,B)}_p :=
\int\limits_{A\le \xi_0 \le \cdots \le \xi_{p-1} \le B} \prod\limits_{j=0}^{p-1} \left(\xi_{j-1} – \xi_j\right) \cdot \prod\limits_{j=0}^{p-1} \frac{d \xi_j}{\xi_j^{2 \beta}} = (-1)^p \frac{\frac{\pi}{2 \beta-2}}{(2\beta-2)^{2 p-1} (p!)^2 \sin(\frac{\pi}{2 \beta-2})} \cdot \\
&&
\sum\limits_{q=0}^p \binom{p}{q} A^{(2-2\beta)(p-q)} B^{(2-2\beta) q-1} \left(A q \binom{p}{q+\frac{1}{2\beta-2}} – B(q-\frac{1}{2\beta-2})\binom{p}{q-\frac{1}{2\beta-2}}\right)
\end{eqnarray}
Again, setting $\beta=0$ we obtain:
\begin{equation}
J_{p}^{(A,B)}\left(0\right) := (-1)^p\frac{(A-B)^{2p}}{(2p)!}
\end{equation}
as it should be.Likewise setting $\beta=2$ we get:
\begin{equation}
J_{p}^{(A,B)}\left(2\right) := (-1)^p\frac{(A-B)^{2p}}{(2p+1)! (A B)^{2 p} B } \left(2p A+B\right)
\end{equation}

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