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Let $M$ be a real symmetric positive definite matrix of size $n \times n$, and let $\log M$ denote its (principal) matrix logarithm.

Is it possible to evaluate the following integral in closed form?

$$Y = \displaystyle\int_{0}^{\infty} \left(M + tI\right)^{-1} (\log M) \left(M + tI\right)^{-1}\:\mathrm{d}t.$$

Not sure if it helps in evaluation, but the above integral has two other representations: $$Y = \displaystyle\int_{-\infty}^{\infty}M^{js}M^{-1/2}(\log M) M^{-1/2} M^{-js} \displaystyle\frac{\pi}{2\cosh^{2}(\pi s)}\:\mathrm{d}s,\\ = \displaystyle\int_{0}^{\infty} \displaystyle\int_{0}^{\infty}\exp(-sM)(\log M)\exp(-tM)\displaystyle\:\frac{\mathrm{d}s\:\mathrm{d}t}{s+t},$$ where $j=\sqrt{-1}$. That the three integral representations are equivalent can be found here (see eqns. (5.4.5), (5.4.8) and the last equation before Appendix with $H=K\equiv M$ and $X\equiv\log M$).

Motivation: the above formulas give the solution of the matrix equation: $\int_{0}^{1}M^{t}YM^{1-t}\mathrm{d}t = \log M$.

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    $\begingroup$ Isn't it just $Y = M^{-1}\log M = (\log M)\,M^{-1}$? $\endgroup$ – Robert Bryant Sep 5 '17 at 11:26
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just go to a basis where $M$ is diagonal, evaluate the integral and find $Y=M^{-1}\log M$.

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First, note that all the factors under the integral commute and $\log(M)$ could be taken out of the integral. Next, note that $\frac{d}{dt} \log(-t) = \frac{(-1)}{t}$ and recall the the holomorphic functional calculus formula for matrices/operators, namely $f^{(n)}(A) = \oint \frac{n! f(t)}{(tI - A)^n} dt$ for any contour the simply and positively encircles the spectrum of $A$. The definite integral can be turned into a contour integral by throwing in a function that has an appropriate branch cut along the positive real axis, which then deforms to positively and simply encircle the spectrum of $(-M)$: \begin{align*} Y &= \log(M) \oint \frac{log(-t)}{-2\pi i} \frac{1}{(tI-(-M))^{1+1}} dt \\ &= \log(M) \frac{(-1)}{(-M)} = \log(M) \frac{1}{M} . \end{align*}

For this calculation, $M$ need not even be symmetric, as long as its spectrum does not intersect the positive real axis.

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