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Define $\mathbf{A}(s)$ as
$$
\mathbf{A}(s) = \int_0^\infty \mathbf{B}(x) e^{-sx} \mathrm{d}x.
$$
Suppose I know $\mathbf{A}(s)$, i.e., I have an explicit formula for it which is valid for all complex numbers $s$. However, I don't know anything about $\mathbf{B}(x)$.

Is it possible to calculate $\mathbf{A}(\mathbf{C})$, defined as
$$
\mathbf{A}(\mathbf{C}) = \int_0^\infty \mathbf{B}(x) e^{\mathbf{-C}x} \mathrm{d}x,
$$
for a given matrix $\mathbf{C}$? If it helps, I know that $\mathbf{C}$ is diagonalizable and its diagonalization is
$$
\mathbf{C} = \mathbf{S} \mathbf{D} \mathbf{S}^{-1},
$$
where $\mathbf{D} = \mathrm{diag}(d_1, ..., d_n)$.

But now
$$
\mathbf{A}(\mathbf{C}) = \left(\int_0^\infty \mathbf{B}(x) \mathbf{S} e^{\mathbf{-D}x} \mathrm{d}x \right) \mathbf{S}^{-1},
$$
wherein the $k$th column of
$$
\int_0^\infty \mathbf{B}(x) \mathbf{S} e^{\mathbf{-D}x} \mathrm{d}x
$$
is the same as the $k$th column of
$$
\int_0^\infty \mathbf{B}(x) \mathbf{S} e^{-d_kx} \mathrm{d}x,
$$
which is the same as
$$
\mathbf{A}(d_k)\mathbf{S}.
$$

Sun, 31 Jan 2021 10:16 GMT