Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am interested in the following function (Mellin transform of matrix exponential): $$\int_0^{\infty} x^{s-1} e^{-A-Bx}d x$$ Where $x$ and $s$ are scalars, but $A$ and $B$ are matrices with $B\succ 0$.

When $A$ and $B$ commute then it is easy to compute: simultaneously diagonalizing gives gamma functions on the diagonal. Are there any other cases in which you can say something about this function? Has it been studied? Is there an analytic continuation?

share|improve this question
1  
One can also compute this when $[A,[A,B]]=0=[B,[A,B]]$, albeit I do not know how to find interesting matrices like this. –  Ralph Furmaniak Oct 20 '11 at 7:46
    
Have you tried working out what it looks like explicitly for some small $A$'s and $B$'s? –  David Loeffler Oct 20 '11 at 8:55
    
Does $B > 0$ meant that $B$ is symmetric positive definite? Is $A$ symmetric also then? Or do you mean something else? –  Igor Rivin Oct 20 '11 at 9:42
    
Yes, $B$ is positive definite to ensure rapid convergence. I'd be happy with some results when $A$ and $B$ are symmetric, but I'd prefer not to assume it. Some non-symmetric cases are doable: When $A+B x=\left[ \begin{array}{cc} 0 & x \\ -1 & 0 \end{array}\right]$ then the function is $2\cos(\pi s)\Gamma(2s) \left[ \begin{array}{cc} 1 & 2s \\ -2s & 1 \end{array}\right]$ –  Ralph Furmaniak Oct 20 '11 at 18:21
add comment

1 Answer

This is not an answer, just an overgrown comment. I hope it suggests something non-boring.

Assume that $-A$ is Hermitian, and $B \ge 0$. Let $G(s)$ denote your matrix valued integral.

Then, we have (am writing $A$ instead of $-A$ to be consistent with some standard notation):

$$g(s) := \mbox{tr}(G(s)) = \int_0^\infty x^{s-1}\mbox{tr}(e^{A-x B})dx.$$

Assuming that the BMV Conjecture is true (see e.g., apparent proof), we can represent the trace term as:

$$\mbox{tr}(e^{A-x B}) = \int_0^\infty e^{-tx}d\mu(t; A,B),$$

where $\mu(; A, B)$ is a positive measure supported by $[0,\infty)$.

Then, assuming that the order of integration can be changed, we have

$$g(s) = \int_0^\infty\int_0^\infty e^{-tx}x^{s-1}dxd\mu(t) = \int_0^\infty \Gamma(s)t^{-s}d\mu(t; A,B),$$ at which point I run out of ideas.

PS: Do you care about just $2 \times 2$ matrices, or then can be $n \times n$?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.