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Let $p$ be a positive integer, let $(q_0,q_1,\cdots,q_p)$ be a sequence of positive integers and let $\beta \neq 1/2$ be a positive number. Then let $B>A>0$. The question is to prove the following identity for a integration over a $p$ dimensional simplex. \begin{eqnarray} &&J^{(A,B)}_{\vec{q},p} := \int\limits_{A\le \xi_0 \le \xi_ \le \cdots \le \xi_{p-1} \le B} \prod\limits_{j=0}^p \left(\xi_{j-1} - \xi_j\right)^{q_j} \cdot \prod\limits_{j=0}^{p-1} \frac{d \xi_j}{\xi_j^{q_{j+1} 2 \beta}}=\\ && (-1)^{Q_P} A^{q_0} \cdot \sum\limits_{m=0}^p \sum\limits_{l=0}^{q_m} \frac{(-1)^{l+q_m} \binom{q_m}{l} }{ \prod\limits_{\stackrel{\eta=0}{\eta\neq m}}^p\left[(q_\eta+1) \binom{l+\eta-m-(1-2\beta)(Q_m-Q_\eta)}{q_\eta+1}\right]} \cdot A^{(1-2\beta)(Q_m-Q_0)+m-l} B^{(1-2\beta)(Q_p-Q_m)+p-m+l} \end{eqnarray} Here $\xi_{-1}=A$ and $\xi_{p}=B$ and $Q_j := q_0+q_1+\cdots+q_j$. The number of terms on the right hand side equals $q_0+q_1+\cdots+q_p+(p+1)$.

Note 1: In the case $q_0=q_1=\cdots=q_p=q$ and $\beta :> \beta/q$ we retrieve the result given in Another mutivariable integral over a simplex .

Note 2: Setting $\beta = 0$ we get \begin{equation} rhs = -\left(A-B\right)^{p + \sum\limits_{\eta=0}^p q_\eta} \cdot \frac{\prod\limits_{\eta=0}^p q_\eta!}{\left(p + \sum\limits_{\eta=0}^p q_\eta\right)!} \end{equation} as it should be.

We also provide a similar integral below where the values of exponents in the denominator are all arbitrary. Let $(q_1,\cdots,q_{p+1})$ be positive integers and let $(\beta_1,\cdots,\beta_p)$ be positive reals numbers. Then we have: \begin{eqnarray} &&I^{(A,B)}_{\vec{q},p} := \int\limits_{A\le \xi_0 \le \xi_ \le \cdots \le \xi_{p-1} \le B} \prod\limits_{j=0}^p \left(\xi_{j-1} - \xi_j\right)^{q_{j+1}} \cdot \prod\limits_{j=0}^{p-1} \frac{d \xi_j}{\xi_j^{\beta_{j+1}}}= (-1)^{Q_{p+1}} \sum\limits_{m=1}^{p+1} \sum\limits_{l=0}^{q_m}\\ && \frac{(-1)^{l+q_m} \binom{q_m}{l} }{\prod\limits_{\stackrel{j=1}{j\neq m}}^{p+1} (q_j+1) \binom{l+j-m+Q_j-Q_m+B_{m-1}-B_{j-1}}{q_j+1} } A^{Q_m-B_{m-1}+m-1-l} B^{Q_{p+1}-Q_m-B_p+ B_{m-1}+p-m+1+l} \end{eqnarray} Here $\xi_{-1}=A$ and $\xi_{p}=B$ and $Q_j := q_1+q_2+\cdots+q_j$ and $B_j := \beta_1+\beta_2+\cdots+\beta_j$. Now we check particular cases of the formula above.

Note 1: Let $q_1=q_2=\cdots=q_{p+1}=0$ and $\beta_j = \beta$ for $j=1,\cdots,p$. Then: \begin{equation} rhs = \frac{\left(A^{1-\beta} - B^{1-\beta}\right)^p}{p! (-1+\beta)^p} \end{equation} as it should be.

Note 2: Now, let $q_1=q_2=\cdots=q_{p+1}=0$ and all the beta values be arbitrary. Then: \begin{equation} rhs = \sum\limits_{m=0}^p \frac{(-1)^m A^{-B_m+m} B^{-(B_p-B_m)+p-m} }{\prod\limits_{j=1}^m (j+ B_{m-j}-B_m) \prod\limits_{j=1}^{p-m}(j+B_m - B_{m+j})} \end{equation}

Note 3: Finally, let $q_1=q_2=\cdots=q_{p+1}=1$ and all the beta values be arbitrary. Then we have: \begin{eqnarray} rhs = (-1)^{p+1}\sum\limits_{l=0}^1 (-1)^{l+1} \sum\limits_{m=0}^p \frac{ A^{-B_m+2 m + 1-l} B^{-(B_p-B_m)+2(p-m)+l} }{\prod\limits_{j=1}^m (2j - l + B_{m-j}-B_m)_{(2)} \prod\limits_{j=1}^{p-m}(2j + l +B_m - B_{m+j})_{(2)}} \end{eqnarray}

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