Derivative of matrix-value function in integral representation

Question

I'm learning about matrices and matrix calculus. In Matrix Monotone Functions and Convexity. In: Introduction to Matrix Analysis and Applications there is written that integral representations of matrices are often helpful with calculating derivaties and there is and example:

I understand the integral representation which follow from spectral theorem and how the derivative is calculated, but I don't understand why we can put $\frac{d}{dt}$ inside the integral. How to prove it formally? Does it somehow follow from Taylor expansion of the inverse?

EDIT

My idea is to put $f(t) = \log{(A + tT)}$ and show that

$$\left| \frac{f(h) - f(0)}{h} - \int_0^\infty (xI+A)^{-1}T(xI+A)^{-1}\right| \to 0$$

By the first formula LHS is equal to

$$\left| \int_0^\infty \frac{(xI+A+hT)^{-1} - (xI+A)^{-1}}{h} - (xI+A)^{-1}T(xI+A)^{-1}\right| = \left| \int_0^\infty \sum_{n=2}^\infty (-h)^{n-1} (xI+A)^{-\frac{1}{2}}\left((xI+A)^{-\frac{1}{2}}T(xI+A)^{-\frac{1}{2}}\right)^n(xI+A)^{-\frac{1}{2}}\right| \le \left| h \int_0^\infty \frac{\left\|T\right\|^2}{\left\|xI+A\right\|^3}\sum_{n=0}^\infty \left(\frac{h\left\|T\right\|}{\left\|xI+A\right\|}\right)^n \right|$$

where I used Taylor expansion and bounded the series by operator norms (assuming they are finite...). Now if $h$ is very small, the series is uniformly convergent and also integral is finite. Is it more or less fine?

Answer 1

"why we can put d/dt inside the integral"? I think that it is just basic definition in matrix calculus plus Leibniz's rule for differentiation under the integral sign in calculus.

1) Let $C(t)$ be a matrix whose $(i,j)$-entry $C_{ij}(t)$ is function of $t$.
Then, $\frac{\mathrm{d}}{\mathrm{d} t} C(t)$ is defined as the matrix whose $(i,j)$-entry is $\frac{\mathrm{d}}{\mathrm{d} t} C_{ij}(t)$.

2) Let $B(x, t)$ be a matrix whose $(i, j)$-entry $B_{ij}(x,t)$ is function of $x$ and $t$.
Then, $\int_0^\infty B(x, t) \mathrm{d} x$ is defined as the matrix whose $(i,j)$-entry is $\int_0^\infty B_{i,j}(x,t) \mathrm{d} x$.

From 1) and 2),
$\frac{\mathrm{d}}{\mathrm{d} t} \int_0^\infty B(x, t) \mathrm{d} x$ is equal to the matrix whose $(i,j)$-entry is
$\frac{\mathrm{d}}{\mathrm{d} t} \int_0^\infty B_{ij}(x, t) \mathrm{d} x = \int_0^\infty \frac{\partial }{\partial t} B_{ij}(x, t) \mathrm{d} x$ according to Leibniz's rule for differentiation under the integral sign,
and hence, $\frac{\mathrm{d}}{\mathrm{d} t} \int_0^\infty B(x, t) \mathrm{d} x = \int_0^\infty \frac{\partial }{\partial t} B(x, t) \mathrm{d} x$.

Answer 2

Proceeding along Barabara's idea:

Let $f(t) = \log (A + tT) $. We have \begin{align} &\left\| \frac{f(h)-f(0)}{h} - \int_0^\infty (xI + A)^{-1}T(xI + A)^{-1} \mathrm{d} x\right\|\\ =\ & \left\| \int_0^\infty \frac{(xI + A)^{-1} - (xI+A+hT)^{-1} }{h} - (xI + A)^{-1}T(xI + A)^{-1} \mathrm{d} x\right\|. \end{align} Denote $B = xI + A$. We have \begin{align} B^{-1} - (B+hT)^{-1} &= B^{-1}[I - B(B+hT)^{-1}] \\ &= B^{-1}[I - (I + hTB^{-1})^{-1}]\\ &= B^{-1}[I - (I + hTB^{-1})^{-1}(I + hTB^{-1} - hTB^{-1})]\\ &= h B^{-1}(I + hTB^{-1})^{-1} TB^{-1}. \end{align} So, we have \begin{align} &\frac{B^{-1} - (B+hT)^{-1}}{h} - B^{-1}TB^{-1}\\ =\ & B^{-1}(I + hTB^{-1})^{-1} TB^{-1} - B^{-1}TB^{-1}\\ =\ & B^{-1}[(I + hTB^{-1})^{-1} - I]TB^{-1}\\ =\ & B^{-1}[(I + hTB^{-1})^{-1}(I + hTB^{-1} - hTB^{-1}) - I]TB^{-1}\\ =\ & - h( B + hT)^{-1}T B^{-1}TB^{-1}. \end{align} So, we have \begin{align} &\left\| \frac{f(h)-f(0)}{h} - \int_0^\infty (xI + A)^{-1}T(xI + A)^{-1} \mathrm{d} x\right\|\\ =\ & |h|\left\|\int_0^\infty (xI + A + hT)^{-1}T (xI + A)^{-1}T(xI + A)^{-1} \mathrm{d} x\right\| \to 0. \end{align} Remark: One need to prove that $\int_0^\infty (xI + A)^{-1}T (xI + A)^{-1}T(xI + A)^{-1} \mathrm{d} x$ is finite. Omitted here.