TITLE: Show that $\log \det(A\cdot B^{-1})=-\int_{0}^\infty tr(e^{-t\cdot A}-e^{-t\cdot B}){dt\over t}$
QUESTION [5 upvotes]: Let $$A,B:V\to V$$ be positive definite operators in complex linear space with inner product $V$ such that $\text{dim}V<\infty$ Show that $$\log \left(\det(A\cdot B^{-1})\right)=-\int_{0}^\infty tr(e^{-t\cdot A}-e^{-t\cdot B}){dt\over t}$$
Could anyone give some hints how to prove this?

REPLY [5 votes]: Nice question. By Frullani's theorem, if $\alpha,\beta >0$, we have:
$$ -\int_{0}^{+\infty}\frac{e^{-\alpha t}-e^{-\beta t}}{t}\,dt = \log(\alpha)-\log(\beta)\tag{1} $$
but the trace of $e^{-tA}$ is just the sum  $\sum_{i=1}^{m} e^{-t\alpha_i}$ where $\alpha_1,\ldots,\alpha_m$ are the eigenvalues of $A$, since:
$$\text{Tr}\,e^{-tA}=\text{Tr}\sum_{n=0}^{+\infty}\frac{(-1)^n t^n}{n!}A^n  =\sum_{n=0}^{+\infty}\frac{(-1)^n t^n}{n!}\text{Tr}(A^n)=\sum_{n=0}^{+\infty}\frac{(-1)^n t^n}{n!}\sum_{i=1}^{m}\alpha_i^n.\tag{2}$$
Given that $A,B$ are positive definite, their eigenvalues are positive, so by $(1)$ and $(2)$ we have:
$$\exp\left(-\text{Tr}\int_{0}^{+\infty}\frac{e^{-A t}-e^{-B t}}{t}\,dt\right)=\frac{\prod_{i=1}^{m}\alpha_i}{\prod_{i=1}^{m}\beta_i}=\frac{\det A}{\det B}=\det(A\cdot B^{-1})\tag{3}$$
as wanted, since by the spectral theorem we may assume without loss of generality that $A$ is a diagonal matrix.