I would like to compute $Y = \int^{\infty}_{-\infty} e^{-itx} (Ix-A)^{-1}dx$, where $A$ is a known square matrix. I am using the semi-circle contour from $-\infty$ to $\infty$. From the Cauchy integral formula, $Y = \oint_{semicircle} - \int^{}_{round part}$.
[EDIT]: as seen on this contour map.
Using $\oint f(z) (Iz-A)^{-1}dz = 2\pi if(A) $ for $|z| > ||A|| $, the round part integral can be evaluated to be $\frac{2\pi if(A)}{2}$ from this contour: contour to evaluate outer semi-circle.
However, I am stuck on computing $\oint_{semicircle}$ which should be $\oint f(z) dz = 2\pi i \sum Res$. How should the residues be computed for matrix contour integrals?
I am aware that the singularities occur at eigenvalues of $A$. However, I am clueless about computing them.
HINT: let any matrix $B$ and let $[B]_{j,k}$ means the coefficient of row $j$ and column $k$, then
$$\left[\int f(x)B(x)\, dx\right]_{j,k}=\int f(x)[B(x)]_{j,k}\, dx$$
So you can evaluate the integral coefficient by coefficient of the matrix $B$. Also remember that
$$G\int f(x)\, dx=\int Gf(x)\, dx$$ for any bounded linear operator $G$ (by example, a square matrix). So you can use the matrix $(Ix-A)^{-1}$ in Jordan form if you please (or any other form easy to manipulate) and after change it back to it standard basis.