Idea for computing integral (similar to beta function) Does anyone have any idea about how to compute the following integral? Or any guess of what it could be?
$$\int_{s<u_1<\cdots <u_p<t} (u_1-s)^{\alpha_1/2}\cdots (u_p-u_{p-1})^{\alpha_p/2}du_1\cdots du_p$$
where $\alpha_1+\cdots+\alpha_p = p$. I was thinking of the Beta function but I do not arrive at anything useful.
Thank you guys
 A: One can compute it in a "beta-function way". I'll suppose $s=0$ (as it depends only on the difference $t-s$). Substitution $x_1=u_1$, $x_2=u_2-u_1$,... ,$x_p=u_p-u_{p-1}$ brings the integral to
$$I(t)=\int_{x_i>0,\sum_i x_i<t}x_1^{\alpha_1/2}\dots x_p^{\alpha_p/2}dx_1\dots dx_p.$$
Clearly
$$I(t)=t^{\sum_i(\alpha_i/2+1)}I(1)$$
and thus
$$dI(t)/dt=\sum_i(\alpha_i/2+1)t^{\sum_i(\alpha_i/2+1)-1}I(1).$$
Let us now compute
$$J:=\int_{x_i>0}x_1^{\alpha_1/2}\dots x_p^{\alpha_p/2}e^{-\sum_i x_i}dx_1\dots dx_p$$
in two ways. On one hand, it is (by definition of $\Gamma$)
$$J=\prod_i\Gamma(\alpha_i/2+1).$$
On the other hand (setting $t=\sum_i x_i$) it is
$$J=\int_0^\infty \frac{dI(t)}{dt} e^{-t}dt=\sum_i(\alpha_i/2+1)I(1)\int_0^\infty t^{\sum_i(\alpha_i/2+1)-1}e^{-t}dt$$
$$=\sum_i(\alpha_i/2+1)I(1)\Gamma(\sum_i(\alpha_i/2+1))=I(1)\Gamma(\sum_i(\alpha_i/2+1)+1).$$
We thus got (modulo mistakes on the way)
$$I(t)=\frac{\prod_i\Gamma(\alpha_i/2+1)}{\Gamma(\sum_i(\alpha_i/2+1)+1)}t^{\sum_i(\alpha_i/2+1)}.$$
