I have $A$ and $B$ square matrices and I want to get rid of the integral in $$ \int_0^Te^{-As}(-A+B)e^{Bs}~ds. $$
This looks very related to the formula $$ \left(\int_0^T e^{At} dt \right) A + I = e^{AT} $$ that was mentioned here . But I don't get it...
I found a closed form solution by means of the theory of differential equations:
If someone gets a direct proof by matrix calculations, I will be happy to accept it. Anyways, here is my solution:
Consider the (matrix-valued) linear ODE: $$ \dot X = BX, \quad X(0)=I $$
and its solution $X(t) = e^{Bt}$.
The same $X$ then solves $$ \dot X = AX + (B-A)e^{Bt}, \quad X(0)=I, $$ which gives $X(t) = e^{At} + \int_0^t e^{A(t-s)}(B-A)e^{Bs}~ds$.
Thus $$ X(t) = e^{Bt} = e^{At} + \int_0^t e^{A(t-s)}(B-A)e^{Bs}~ds = e^{At} + e^{At}\int_0^t e^{-As}(B-A)e^{Bs}~ds $$ from where finds that $$ \int_0^t e^{-As}(B-A)e^{Bs}~ds = e^{-At}(e^{Bt}-e^{At}) $$ EDIT: I have removed an identity in the last formula that only holds for commuting matrices as @loupblanc pointed out. In general it only holds that $$ \int_0^t e^{-As}(B-A)e^{Bs}~ds = e^{-At}(e^{Bt}-e^{At}) = e^{-At}e^{Bt}-I. $$
We agree on the fact that the product of square matrices usually has no property of commutation which makes the thing difficult. Still you know that by definition: $$e^{-As}=\sum_{k=0}^{\infty}A^k\frac{(-s)^k}{k!}$$ And : $$e^{Bs}=\sum_{i=0}^{\infty}B^i\frac{s^i}{i!}$$ Thus : $$\int_{0}^{T}e^{-As}(-A+B)e^{Bs}ds=\int_{0}^{T}\sum_{k=0}^{\infty}A^k\frac{(-s)^k}{k!}(-A+B)\sum_{i=0}^{\infty}B^i\frac{s^i}{i!}ds$$ $$=\int_{0}^{T}\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^k(-A+B)B^i\frac{(-1)^ks^{i+k}}{k!i!}ds$$ Then you can use an interversion theorem to get the integral into the sum and to write : $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^k(-A+B)B^i\frac{(-1)^k}{k!i!}\int_{0}^{T}s^{i+k}ds$$ You can certainly go back to an expression with exponentials from there. $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^k(-A+B)B^i\frac{(-1)^k}{k!i!}\frac{T^{i+k+1}}{i+k+1}$$ $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^{k+1}B^i\frac{(-1)^{k+1}}{k!i!}\frac{T^{i+k+1}}{i+k+1}+\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^kB^{i+1}\frac{(-1)^k}{k!i!}\frac{T^{i+k+1}}{i+k+1}$$ $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}\frac{(-TA)^{k+1}}{k!}\frac{(TB)^i}{i!}\frac{1}{i+k+1}+\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}\frac{(-TA)^{k}}{k!}\frac{(TB)^{i+1}}{i!}\frac{1}{i+k+1}$$
