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I am wondering how to integrate the function

$x \mapsto \operatorname{tr}\bigl\{(\mathbf{A}x+\mathbf{B})^{-1}\mathbf{C}\bigr\}$

In my case, the matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ are (strictly) positive definite. If $\mathbf{C} = \mathbf{A}$ it is easy to see that

$\ln\det(\mathbf{A}x+\mathbf{B})$

is an antiderivative. But what if $\mathbf{C} \neq \mathbf{A}$?

Thanks,
jens

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    The account used to post the question here was unregistered, and it appears you were unable to get back into the account. Eric Naslund did some mod voodoo. Anyway: log is the inverse of exp and has a power series expansion equivalent to the real case (with convergence technicalities, naturally). There are also very interesting and deep issues dealing with noncommutativity relevant to Lie theory (see the BCH formula). If you plug in all real values you should get a real trace, otherwise I'm not sure what I can say.2012-07-01

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By the linearity of integration, trace, and matrix multiplication, we can pull $\int$ inside:

$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx=\mathrm{tr}\left( \int (Ax+B)^{-1}dx\, C\right).$

I originally thought $\log(Ax+B)\;A^{-1}$ would be the antiderivative for this, but oenamen helpfully points out this can't be said in full generality ($\log XY=\log X+\log Y$ does not necessarily hold if the matrices $X$ and $Y$ don't commute!); instead, we need a special $B^{-1}$ factor. We have:

$\frac{d}{dx}\log(I+xU) =\quad U(I+xU)^{-1} =(I+xU)^{-1}U.$

Hence, applying the above alongside $X^{-1}Y^{-1}=(YX)^{-1}$:

$\begin{array}{c l} \frac{d}{dx}\log \big(B^{-1}(Ax+B)\big) & =\frac{d}{dx}\log(I+xB^{-1}A) \\ & = (I+xB^{-1}A)^{-1}B^{-1}A \\ & = \big(B\cdot(I+B^{-1}Ax)\big)^{-1}A \\ & = (Ax+B)^{-1}A. \end{array}$

Therefore we conclude

$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx =\mathrm{tr}\left(\log\big(I+xB^{-1}A)\; A^{-1}C\right)+const. $

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    Ah, for a second there I thought you had some superuser privileges!2012-07-01