Integral of a determinant is bounded for each dimension The motivation for this problem is bounding a certain probability related to a multivariate normal distribution, although my question is in the realm of calculus and linear algebra.
For $x_1<...<x_p \in [0,1]$, define the matrix $B(x_1,...,x_p) = [\min(x_i,x_j)]_{1\le i,j\le p}$. $B$ is a symmetric matrix with a fairly simple structure. Consider the real valued function $$D(x_1,...,x_p)= det( B(x_1,...,x_p))^{-1/2}$$.
What I would like to show is that
$$
\int_0^1 \int_0^{x_p} \int_0^{x_{p-1}} \cdots \int_0^{x_2} D(x_1,...,x_p) dx_1 ... dx_p < \infty. 
$$
Clearly this holds for $p=1$, in which case the integral reduces to $\int_0^1 x_1^{-1/2}dx_1 =2$. Brute force shows that it also holds for $p=2$. I'm wondering if one of you bright folks could find a simple argument to show it holds for all $p$.
 A: I will re-enumerate the variables as follows
$$0< y_p<y_{p-1}<\ldots < y_1<1=y_0$$
The determinant can be evaluated explicitly as
$$\det B(y_p,y_{p-1},\ldots, y_1) =y_p(y_{p-1}-y_p)\ldots (y_1-y_2)$$
The integral takes the form
$$
\int\limits_0^1 \int\limits_0^{y_1} \int\limits_0^{y_2} \cdots \int\limits_0^{y_{p-1}} [y_p(y_{p-1}-y_p)\ldots (y_1-y_2)]^{-1/2} dy_p ... dy_1 
$$
Let $$\hspace{2cm} y_k=t_ky_{k-1},\quad 0<t_k<1,\ 1\le k\le p$$  Then $$y_k=t_kt_{k-1}\ldots t_1,\quad 1\le k\le p$$
The matrix of the change of  variables from $y_k$-s to $t_k$-s is triangular and its determinant is equal
$$ t_1^{p-1}t_2^{p-2}\ldots t_{p-1}$$
Observe that
$$ y_p(y_{p-1}-y_p)\ldots (y_1-y_2)=t_1^pt_2^{p-1}\ldots t_p(1-t_p)\ldots (1-t_2)$$
Thus the integral will be transformed to
$$ \int\limits_0^1 \int\limits_0^{1} \int\limits_0^{1} \cdots \int\limits_0^{1} (t_1^pt_2^{p-1}\ldots t_p)^{-1/2}\,[(1-t_2)\ldots (1-t_p)]^{-1/2}\\ t_1^{p-1}t_2^{p-2}\ldots t_{p-1}\,\,dt_1 ... \,dt_p$$
The variables are separated hence  we obtain the product of one variable integrals (for $k\ge 2$)
$$\int\limits_0^1 (t_k^{p-k+1})^{-1/2}t_k^{p-k}(1-t_k)^{-1/2}\,dt_k\\ = \int\limits_0^1 t_k^{(p-k-1)/2} (1-t_k)^{-1/2}\,dt_k={\Gamma({1\over 2})\Gamma({p-k+1\over 2})\over \Gamma({p-k+2\over 2})}$$ and
$$\int\limits_0^1 t_1^{p/2-1}\,dt_1={2\over p+2} $$
The final result is the product of one variable integrals:
$${\pi^{p/2}\over \Gamma({p\over 2}+1)}$$
