$T$ is a positive selfadjoint densely defined on $H$ (Hilbert), only has point spectrum $\sigma (T)=\{\lambda_k \}_{k\in\mathbb{N}}.\lambda_k < \lambda_{k+1}\forall k$ and $\{u_k\}_{k\in \mathbb{N}}$ is the respective orthonormal base of eigenvectors of $T$.
$Tr(A)=\sum_n <u_n,Au_n>$ for $A$ a positive operator on $H$. If $Tr(A)<\infty$ then $A$ is trace class. Assume $e^{-aT}$ is trace class where $a>0$.
$A,B$ are positive bounded operators on $H$.
I wanna show $\int_0^1 Tr(e^{-a(1-t)T}A^{∗}e^{-atT}B) dt=\sum_{m,n;m\neq n}<u_m,A^{∗}u_n><u_m,Bu_n>\frac{e^{-a\lambda_m}-e^{-a\lambda_n}}{-a(\lambda_m-\lambda_n)}+\sum_n<u_n,A^{∗}u_n><u_n,Bu_n>e^{-a\lambda_n}$
I know that the trace is independent of the base, its ciclicity, but I can't get a double summatory nor the product of inner products.
I got $...=\int_0^1\sum_n e^{-a(1-t)\lambda_n}<Au_n,e^{-atT}Bu_n>dt$
Any help is regarded.
Observe \begin{align} \operatorname{Tr}(e^{-a(1-t)T}A^\ast e^{-atT}B) =&\ \sum_n \langle Ae^{-a(1-t)T}u_n, e^{-atT}Bu_n\rangle\\ =&\ \sum_{n, m}\langle Ae^{-a(1-t)T}u_n, u_m\rangle \langle u_m, e^{-atT}Bu_n\rangle\\ =&\ \sum_{n, m}e^{-a(1-t)\lambda_n}\langle Au_n, u_m\rangle \langle u_m, e^{-atT}Bu_n\rangle. \end{align} Summing up the diagonal terms yields \begin{align} \sum_n e^{-a(1-t)\lambda_n}\langle Au_n, u_n\rangle \langle u_n, e^{-atT}Bu_n\rangle =&\ \sum_n e^{-a(1-t)\lambda_n}\langle Au_n, u_n\rangle \langle e^{-atT}u_n, Bu_n\rangle\\ =&\ \sum_n e^{-a\lambda_n}\langle u_n, A^\ast u_n\rangle \langle u_n, Bu_n\rangle. \end{align} For the off-diagonal terms, we have \begin{align} \sum_{n\ne m}e^{-a(1-t)\lambda_n}\langle Au_n, u_m\rangle \langle u_m, e^{-atT}Bu_n\rangle =&\ \sum_{n\ne m}e^{-a(1-t)\lambda_n-at\lambda_m}\langle u_n, A^\ast u_m\rangle \langle u_m, Bu_n\rangle. \end{align}
I think the rest should be self explanatory.