Correspondence between two matrices Suppose $B$ is a positive definite matrix with determinant $1 $ and  $$ A = \frac{1}{2} \int_0^\infty \frac{(B+sI)^{-1}}{\sqrt{\mbox{det}(B+sI)}} ds $$
Then, how does one prove that this provides a one to one onto correspondence between positive definite matrices $B$ with determinant $1$ and positive definite matrices $A$ with trace $1$. 
Thank you very much. 
 A: Without loss of generality, let $B$ be diagonal matrix with diagonal elements $b_i$. Consider
$$ \begin{eqnarray}
  A_{11} &=& \frac{1}{2} \int_0^\infty \frac{1}{(s+b_1)^{3/2}} \prod_{k=2}^n \frac{1}{(s+b_k)^{1/2}} \mathrm{d} s \\  
  &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} s \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n x_1^{1/2} x_2^{-1/2}\cdots x_n^{-1/2} \mathrm{e}^{-(b_1+s)x_1} \cdots \mathrm{e}^{-(b_n+s)x_n} \\
  &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \frac{x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2}}{x_1 + x_2 + \ldots + x_n} x_1 \mathrm{e}^{-\sum_i b_i x_i}
\end{eqnarray}
$$
Notice that integral representation for the sum $\sum_i A_{ii}$ will be:
$$ \begin{eqnarray}
  \operatorname{Tr}(A) &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \frac{x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2}}{x_1 + x_2 + \ldots + x_n} \left( \sum_i x_i \right) \mathrm{e}^{-\sum_i b_i x_i} \\
  &=& \frac{1}{ \pi^{n/2}} \int_0^\infty \mathrm{d} x_1 \cdots \int_0^\infty \mathrm{d} x_n \left(x_1^{-1/2} x_2^{-1/2}\cdots x_n^{-1/2} \right) \mathrm{e}^{-\sum_i b_i x_i} = \frac{1}{\sqrt{\det B}} = 1
\end{eqnarray}
$$
