A certain multidimensional integral. Consider a following multidimensional integral:
\begin{equation}
\bar{I}^{(t_0,t)}_p := \int\limits_{t_0 \le \xi_0 \le \cdots \le \xi_{p-1} \le t} 
\prod\limits_{j=0}^p (\xi_{j-1}-\xi_j) \cdot \prod\limits_{j=0}^{p-1} \frac{d \xi_j}{\xi_j^2}
\end{equation}
subject to $\xi_{-1}=t_0$ and $\xi_p=t$.
Now, by using elementary integration and mathematical induction we have shown that:
\begin{equation}
\bar{I}^{(t_0,t)}_p = \sum\limits_{j=0}^p \frac{1}{j!} \binom{2 p-j}{p-j} \cdot \left( t_0 - (-1)^j t \right) \cdot \left( \log(t/t_0)\right)^j 
\end{equation}
On the other hand the integral in question is a special case of a more general integral given in Dyson-expansion like multidimensional integral . Therefore the question would be how can we show that the right hand side of the equation in the link provided is equal to the right hand side of our equation above.
In other words we want to show that:
\begin{equation}
\bar{I}^{(t_0,t)}_p  = \lim\limits_{\beta \rightarrow 1}\sum\limits_{m=0}^p \sum\limits_{l=0}^1 (-1)^{l+p+2} \frac{\binom{1}{l}}{2^p \prod\limits_{\stackrel{j=0}{j\neq p-m}}^p  \binom{l+(2 \beta-2) \cdot (p-m-j)}{2}} \cdot t^{l+(2-2 \beta)m} \cdot t_0^{1-l+(2-2 \beta)(p-m+1)}
\end{equation}
The right hand side of the last equation above is just equal to the right hand side in the link provided taken at $p_j=1$ for all $j=0,\cdots,p$.
 A: Consider the right hand side of the last equation in the formulation of the question. By substituting $\beta = 1+\epsilon$ and then simplifying the result we obtain the following form:
\begin{equation}
rhs = \frac{1}{2^{2 n}}\sum\limits_{l=0}^1 \sum\limits_{m=0}^p (-1)^{l+p+2}
\frac{\binom{1}{l}}{[m! (p-m)!]^2}  \cdot \left\{
\frac{   \exp (-\epsilon (2 (-m+p+1) \log (t_0)+2 m \log (t)))}{ \epsilon^p \prod\limits_{j=1}^m \left(\epsilon+\frac{(-1)^l}{2 j}\right)  \prod\limits_{j=1}^{p-m} \left(\epsilon-\frac{(-1)^l}{2j}\right)}
\right\}
\cdot
t^l t_0^{1-l} 
\end{equation}
Clearly the expression in curly brackets has a pole in $\epsilon$ of order $p$. Therefore in order to take the limit $\epsilon \rightarrow 0$ all we need to do is to extract the appropriate coefficient of the Laurent expansion about zero. In other words we replace:
\begin{equation}
\left\{ \cdot \right\} \longrightarrow \frac{1}{p!} \frac{d^p}{d \epsilon^p} \left( \epsilon^p \left\{ \cdot \right\} \right)
\end{equation}
The denominator is a polynomial in $\epsilon$ of order $p$ whose roots we know.Therefore we can expand the inverse denominator into simple fractions by using the following formula:
\begin{equation}
\prod\limits_{j=1}^m \frac{1}{(x + \frac{1}{2 j})} \cdot
\prod\limits_{j=1}^{p-m} \frac{1}{(x - \frac{1}{2 j})}
=
\sum\limits_{j=1}^m \frac{(-2 j)^{p-1} (-1)^{j-1}}{(x + \frac{1}{2 j})}  \frac{\binom{m}{j}}{\binom{p-m+j}{j}}+
\sum\limits_{j=1}^{p-m} \frac{(+2 j)^{p-1} (-1)^{j-1}}{(x - \frac{1}{2 j})}  \frac{\binom{p-m}{j}}{\binom{m+j}{j}}+
\end{equation} 
where $x := (-1)^l \epsilon$. Having done this all we need to do is to use the Leibnitz formula for chain differentiation end we arrive at the following result:
\begin{eqnarray}
&&rhs = \sum\limits_{l=0}^1 \sum\limits_{m=0}^p \frac{(-1)^{p+1}}{[(p-m)! m!]^2} \cdot
\sum\limits_{k=0}^p
\left\{\right.\\
&&\left.
\sum\limits_{j=1}^m \frac{(-1)^{l k+j-1}}{k!} \left( m \log(t) + (p-m+1) \log(t_0)\right)^k j^{2 p-k} \frac{\binom{m}{j}}{\binom{p-m+j}{j}}
+\right.\\
&&\left.
\sum\limits_{j=1}^{p-m} \frac{(-1)^{(l+1) k+j-1}}{k!} \left( m \log(t) + (p-m+1) \log(t_0)\right)^k j^{2 p-k} \frac{\binom{p-m}{j}}{\binom{m+j}{j}}
\right. \\
&&\left.
\right\}
(-1)^l t^l t_0^{1-l}
\end{eqnarray}
The expression still needs to be simplified but it is already readily seen that we are dealing with something that is a polynomial of order $p$ in the $\log$ and of order one in the times $t$ and $t_0$. Now, by expanding the parentheses involving the logarithms in a binomial series and then collecting all terms that involve a given power of $\log(t/t_0)$  we easily obtain the sought after relationship if we only notice that the following identity holds:
\begin{eqnarray}
&&\sum\limits_{m=0}^p \frac{(-1)^{p+1}}{[(p-m)! m!]^2} \cdot m^k \cdot \left( \sum\limits_{j=1}^m (-1)^{j-1} j^{2 p-k_1} \frac{\binom{m}{j}}{\binom{p-m+j}{j}} + \sum\limits_{j=1}^{p-m} (-1)^{j-1} (-j)^{2 p-k_1} \frac{\binom{p-m}{j}}{\binom{m+j}{j}}\right)\\
&& = \delta_{k_1,k} \cdot \binom{2 p-k}{p-k}
\end{eqnarray}
valid for $p=1,2,3,\cdots$ and $k=0,\cdots,p$ and $k_1=k,\cdots,p$. Here $0^k = \delta_{0,k}$. It would be nice to prove this identity..
