I am trying to evaluate the following integral which arises from a probability problem $$\iiiint\limits_{\Omega} (1-e^{-(u+v+w)})^{n-4}e^{-(u+v+w)-(n-3)x}d\Omega,$$ where $\Omega$ is the region defined by $0<u\leq v\leq w\leq x<\infty$ and $n$ is an integer $\geq 4$.
An answer using Beta function $B(x,y)=\int_{0}^1 t^{x-1}(1-t)^{y-1}dt$ would be fine (and nice to obtain).
I managed to calculate the analogous integral (albeit multiplied by a constant) in two or three dimensions and got a linear combination of Beta functions, namely $$n(n-1)\iint\limits_{0<u\leq v<\infty} (1-e^{-u})^{n-2}e^{-u-(n-1)v}dudv=nB(n-1,n)$$ quite straightforwardly and $$n(n-1)(n-2)\iiint\limits_{V} (1-e^{-(u+v)})^{n-3}e^{-(u+v)-(n-2)w}dV $$ where $V=0<u\leq v\leq w<\infty$, which equals $$\frac{n(n-1)}{n-2}\left(\frac{1}{2} B\left(n-1,\frac{n-2}{2}\right)-B\left(n-1,n-2\right) \right)$$ (calculated by changing the order of integration).
However, I must say I can't quite do the same with this integral in 4D and any help is appreciated. Even math software like Maple has let me down (it had already on lower dimensions however).
Order changes: first $(y,v)$ omitting other term. $\int_0^wdv\int_{v+w}^{2v+w}dy=\int_w^{2w}dy\int_{\frac{y-w}{2}}^{y-w}dv+\int_{2w}^{3w}dy\int_{\frac{y-w}{2}}^wdv=\int_w^{2w}\frac{y-w}{2}dy+\int_{2w}^{3w}\frac{3w-y}{2}dy$
Next interchange $(y,w)$ we have $\int_0^xdw\int_w^{2w}\frac{y-w}{2}dy=\int_0^xdy\int_\frac{y}{2}^y\frac{y-w}{2}dw+\int_x^{2x}dy\int_\frac{y}{2}^x\frac{y-w}{2}dw$
and $\int_0^xdw\int_{2w}^{3w}\frac{3w-y}{2}dy=\int_0^{2x}dy\int_\frac{y}{3}^\frac{y}{2}\frac{3w-y}{2}dw+\int_{2x}^{3x}dy\int_\frac{y}{3}^x\frac{3w-y}{2}dw$.
I'll leave the $w$ integrations (4 terms) to you.