Daletskii-S.Krein formula proof I've came across to the following equation, known as Daletskii-S.Krein formula. Consider a sufficiently smooth function $h : \mathbb{R} \rightarrow \mathbb{R}$, and let $\mathbf{A}_t = \mathbf{A} + t\mathbf{T}$, where $t = [0,1]$, and both $\mathbf{A}$ and $\mathbf{T}$  are self adjoint operators in a hilbert space $\mathfrak{H}$. Then 
$$
h(\mathbf{A}_t)-h(\mathbf{A}) = t\int\limits_{\sigma(\mathbf{A})}\int\limits_{\sigma(\mathbf{A}_t)} \phi(\lambda,\mu) dE^{\mathbf{A}}\mathbf{T}dE^{\mathbf{A}_t}
$$
or when taking $t \rightarrow 0$,
$$
\left. \dfrac{d h(\mathbf{A}_t)}{t}\right|_{t=0} = \int\limits_{\sigma(\mathbf{A})}\int\limits_{\sigma(\mathbf{A})} \phi(\lambda,\mu) dE^{\mathbf{A}}\mathbf{T}dE^{\mathbf{A}}
$$
My question is how to prove the equation above. I found this equation in 
scienceDirect paper mentioned without proof. The main reference of this formula (ref 18) is difficult to find right now. Any help will be appreciated.
Cheers.
 A: The proof is quite long, see Proposition 6.11 in https://www.sciencedirect.com/science/article/pii/S0022123603003860.
See also section 8.3 of the survey by Birman and Solomyak enter link description here. 
A: As remarked in user92646's answer, the proof is quite involved, but as usual, everything is a lot easier if $h$ is a polynomial. First note that it suffices to consider $h(\lambda)=\lambda^{n+1}$. In this case, $\phi(\lambda,\mu)=\sum_{j+k=n}\lambda^j \mu^k$.
If the integrand $\phi$ is of the form $\phi(\lambda,\mu)=f(\lambda)g(\mu)$, then the double operator integral is simply
$$
\int_{\sigma(X)}\int_{\sigma(Y)}f(\lambda)g(\mu)\,dE^X(\lambda)TdE^Y(\mu)=f(X)Tg(Y).
$$
For $\phi$ as above, this gives
$$
\int_{\sigma(A)}\int_{\sigma(A_t)}\phi(\lambda,\mu)\,dE^{A}TdE^{A_t}=\sum_{j+k=n}A^jT(A+tT)^k.
$$
On the other hand, one can show by induction that
$$
(A+tT)^{n+1}-A^{n+1}=t\sum_{j+k=n}A^j T(A+tT)^k.
$$
When $h$ is not necessarily a polynomial, one can still approximate it by polynomials and $\phi$ by linear combinations of elementary tensors. the hard work is then to show that everything (especially on the right-hand side) converges nicely so that the equation remains valid.
