0
$\begingroup$

I have a more or less stupid problem with an asymptotic value of an integral. consider

$\int_{\infty}^{Q} dx$ $\exp[-4x-a/x] \int_{\infty}^{x} dy$ $\exp[-2y -a/y]$ in the limit $Q \rightarrow \infty.$

I get the following:

$\int_{\infty}^{Q} dx$ $\exp[-4x-a/x] (-1/2)\exp[-2x] \rightarrow (1/2) (1/6) \exp[-6Q]=(1/12) \exp[-6Q]$

however, when I evaluate that integral with Mathematica and compare the asymptotics, it looks like it's falling like $(1/24)\exp[-6Q]$ (there seems to be a factor of 2 difference). where am I making a mistake?

thanks for help. regina

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Have you tried binary search? Pick a formula on the middle of your page, evaluate it numerically and see whether it is off by a factor of 2 or not. Then you know whether the error is before or after that formula. Repeat until the error is found. $\endgroup$
    – David E Speyer
    Commented Jun 2, 2010 at 12:26
  • $\begingroup$ for $a=0$ the factor $1/12$ is certainly correct, and it seems obvious that the large-$Q$ asymptotics cannot depend on the value of $a$, although for large $a$ it may be difficult to reach the asymptotic regime numerically; so my best bet is that your numerics is misleading, and indeed, the $1/12$ coefficient is correct. $\endgroup$
    – Carlo Beenakker
    Commented Jun 27, 2013 at 17:33
  • $\begingroup$ Maybe the error is in what you did with Mathematica. But unless you explain what you did, nobody can tell. $\endgroup$
    – Michael Renardy
    Commented Jun 27, 2013 at 18:56
  • $\begingroup$ This question appears to be off-topic because it lacks motivation and context. If the question is in fact about research level mathematics within the scope defined in the help center, some motivation could clarify this. If it is not, then Math.StackExchange might be a better venue. $\endgroup$
    – Theo Johnson-Freyd
    Commented Dec 15, 2013 at 6:52

1 Answer 1

Reset to default
2
$\begingroup$

this question has been open for half a year, so just to get it off the "unanswered" list, here is some feedback:

for $a=0$ the coefficient $1/12$ is certainly correct, and it seems obvious that the large-$Q$ asymptotics cannot depend on the value of $a$, although for large $a$ it may be difficult to reach the asymptotic regime numerically.

here is the Mathematica output for $a=1$; as you can see, it converges, slowly, to the expected value of $1/12$; the $1/24$ value you quote is certainly far off.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .