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}



You must log in to answer this question.

Browse other questions tagged .