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...
On
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. $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^k(-A+B)B^i\frac{(-1)^k}{k!i!}\frac{T^{i+k+1}}{i+k+1}$$ $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^{k+1}B^i\frac{(-1)^{k+1}}{k!i!}\frac{T^{i+k+1}}{i+k+1}+\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}A^kB^{i+1}\frac{(-1)^k}{k!i!}\frac{T^{i+k+1}}{i+k+1}$$ $$=\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}\frac{(-TA)^{k+1}}{k!}\frac{(TB)^i}{i!}\frac{1}{i+k+1}+\sum_{k=0}^{\infty}\sum_{i=0}^{\infty}\frac{(-TA)^{k}}{k!}\frac{(TB)^{i+1}}{i!}\frac{1}{i+k+1}$$
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 $$ \int_0^t e^{-As}(B-A)e^{Bs}~ds = e^{-At}(e^{Bt}-e^{At}) = e^{-At}e^{Bt}-I. $$