Let $N\ge 1$ be an integer and $\vec{p}:=\left\{p_i\right\}_{i=1}^N$ be nonnegative integers. In addition let $T=0,1$ and $z\in {\mathbb R}_+$. Consider a following multivariate integrals: \begin{eqnarray} {\mathfrak I}_<^{(N,T,\vec{p})}(z) &:=& \int\limits_{0 < \nu_1< \cdots < \nu_N < z} \left[\prod\limits_{l=1}^N \nu_l^{\frac{T}{2}+p_l} e^{-\nu_l}\right] \cdot \prod\limits_{l=1}^N d\nu_l \\ {\mathfrak I}_>^{(N,T,\vec{p})}(z) &:=& \int\limits_{z < \nu_1< \cdots < \nu_N < \infty} \left[\prod\limits_{l=1}^N \nu_l^{\frac{T}{2}+p_l} e^{-\nu_l}\right] \cdot \prod\limits_{l=1}^N d\nu_l \end{eqnarray} Now we have computed the result for $N=2$. The first integral reads: \begin{eqnarray} {\mathfrak I}_<^{(2,T,\vec{p})}(z) = \left\{ \begin{array}{rr} (-1)^{p_1+p_2} \left.\frac{\partial^{p_1+1}}{\partial A_1^{p_1+1}} \frac{\partial^{p_2+1}}{\partial A_2^{p_2+1}} \left( \frac{2 \arctan(\sqrt{\frac{A_1}{A_2}}) - 4\pi {\mathbb T}(\sqrt{2 A_2 z},\sqrt{\frac{A_1}{A_2}})}{\sqrt{A_1 A_2}}\right)\right|_{A_1=1,A_2=1} & \mbox{if $T=1$} \\ (-1)^{p_1+p_2} \left.\frac{\partial^{p_1}}{\partial A_1^{p_1}} \frac{\partial^{p_2}}{\partial A_2^{p_2}}\left( \frac{1}{A_2(A_1+A_2)}- \frac{e^{-A_2 z}}{A_1 A_2}+\frac{e^{-(A_1+A_2)z}}{A_1(A_1+A_2)}\right) \right|_{A_1=1,A_2=1} & \mbox{if $T=0$} \end{array} \right. \end{eqnarray} and the second one reads: \begin{eqnarray} &&{\mathfrak I}_>^{(2,T,\vec{p})}(z) = \\ &&\left\{ \begin{array}{rr} (-1)^{p_1+p_2} \left.\frac{\partial^{p_1+1}}{\partial A_1^{p_1+1}} \frac{\partial^{p_2+1}}{\partial A_2^{p_2+1}} \left( \frac{4\pi {\mathbb T}(\sqrt{2 A_2 z},\sqrt{\frac{A_1}{A_2}})-\pi Erf(\sqrt{A_1 z})Erfc(\sqrt{A_2 z})}{\sqrt{A_1 A_2}}\right)\right|_{A_1=1,A_2=1} & \mbox{if $T=1$} \\ (-1)^{p_1+p_2} \left.\frac{\partial^{p_1}}{\partial A_1^{p_1}} \frac{\partial^{p_2}}{\partial A_2^{p_2}}\left( \frac{e^{-(A_1+A_2)z}}{A_2(A_1+A_2)}\right) \right|_{A_1=1,A_2=1} & \mbox{if $T=0$} \end{array} \right. \end{eqnarray} Here ${\mathbb T}(\cdot,\cdot)$ is the Owen's T-function https://en.wikipedia.org/wiki/Owen%27s_T_function . Now, for every fixed $p_1$ and $p_2$ we can compute the derivatives above using the chain rule and we get a result involving powers of $z$ and exponentials of $z$ in case $T=0$ and the former along with $Erf((\cdot) z)$ in case of $T=1$.
My question is twofold. Firstly is it possible to get a closed form for the derivatives in question? Secondly can we derive the result for $N > 2$ as well ?
Here I provide the answer for $T=0$ and arbitrary $N \ge 2$. We have: \begin{eqnarray} {\mathfrak J}_<^{(N,0,\vec{p})}(z) &=& (-1)^{\sum\limits_{\xi=1}^N p_\xi} \left(\prod\limits_{\xi=1}^N \partial^{p_\xi}_{A_\xi}\right)\left.\left[ \sum\limits_{j=0}^N (-1)^j \frac{\exp(-z \sum\limits_{\xi=n-j+1}^n A_\xi)}{\prod\limits_{l=1}^j (\sum\limits_{\xi=n-j+1}^{n-j+l} A_\xi) \cdot \prod\limits_{l=1}^{n-j} (\sum\limits_{\xi=n-j-l+1}^{n-j} A_\xi)} \right]\right|_{\vec{A}=\vec{1}}\\ {\mathfrak J}_<^{(N,0,\vec{p})}(z) &=& (-1)^{\sum\limits_{\xi=1}^N p_\xi} \left(\prod\limits_{\xi=1}^N \partial^{p_\xi}_{A_\xi}\right)\left.\left[ \frac{\exp(-z\sum\limits_{\xi=1}^n A_\xi)}{\prod\limits_{j=1}^n (\sum\limits_{\xi=n-j+1}^n A_\xi)} \right]\right|_{\vec{A}=\vec{1}}\\ \end{eqnarray}
Unfortunately if $T=1$ the result is much more complicated and to the best of my knowledge cannot be in general expressed through elementary functions.