MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I have a function $X(t)$ whose Laplace transform $\hat{X}(s)$ has a unique pole of largest real part $x_0$, which is a real number. I am able to show that for each $t$, $X(t)$ is a convergent sum of the residues of $\hat{X}(s)e^{st}$, over its poles. If $x_0$ is a pole of order $m$, is it true in general that $X(t)/(t^{m-1}e^{x_0t})$ converges as $t\rightarrow\infty$?

share|cite|improve this question
Any particular reason you're considering this function? – David Roberts Feb 14 '13 at 0:41
Since Eric seems to be AWOL, I will answer on his behalf because I know the background to his question (and indeed suggested he post it). Eric has written down a delay-difference equation related to counting paths by length in a graph where the edges are assigned incommensurate lengths. He has the explicit Laplace transform of the solution to the DDE, and is trying to recover properties of the solution. The issue is that the Laplace transform has a pole at some x0 with all other poles having strictly smaller real part (but the vertical strip over $(x_0-\epsilon,x_0)$ has inf many poles) – Anthony Quas Feb 14 '13 at 19:46
up vote 1 down vote accepted

Yes, this is true. Just write the Laplace transform as a sum of $a/(x-x_0)^m$ and the rest. The inverse Laplace of the first summand is your asymptotic term, and the inverse Laplace transform of the rest is small. (I suppose that your expression "unique pole of the largest real part" means that the real parts of the other poles are at most $x_0-\epsilon$).

share|cite|improve this answer
@Alexandre: My understanding from discussing with Eric is that he anticipates having a pole at $x_0$, and then infinitely many poles in the strip $\{z\colon x_0-\epsilon < \Re(z) < x_0\}$ (but only the single pole on the line $\Re(z)=x_0$). The issue he was facing was that it's easy to see the contribution from any single pole is dominated by the contribution from $x_0$. But how about the combined contribution? – Anthony Quas Feb 14 '13 at 19:41
@Anthony: The combined contribution is usually easy to estimate, but if you need details, you have to state the question with details: you say "convergent sum of residues" How it is convergent? Absolutely? uniformly? Etc. – Alexandre Eremenko Feb 15 '13 at 0:26
The convergence is pointwise: there is a sequence of rectangular contours $\Gamma_n$ that expand to fill the region containing the poles of $\hat{X}(s)$, such that for each fixed value of $t$, $X(t) = \lim_{n\to\infty}\sum_{s_j \in\Gamma_n}\textrm{Res}(\hat{X}(s)e^{st},s_j)$, where $s_j \in \Gamma_n$ denotes a pole $s_j$ in the interior of the contour $\Gamma_n$. – Eric Foxall Feb 15 '13 at 18:08
For a bit more detail, we have that $\hat{X}(s) = (I-B(s))^{-1}\frac{1}{s}$ is a vector-valued function obtained via the matrix $B(s) = a_{ij}e^{-l_{ij}s}$, where each $a_{ij}\geq 0$ and each $l_{ij}> 0$. Poles appear when $\det(I-B(s))=0$. – Eric Foxall Feb 15 '13 at 18:11

Your Answer


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.