Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,549 answers , 19,357 comments
1,470 users with positive rep
410 active unimported users
More ...

Steepest descent for Mellin-type integration

+ 2 like - 0 dislike
20 views

Here I would like to see the behavior of a function as an integral when its argument (which is a parameter in the integral) goes to zero. If I try to evaluate an integral $\int^{i\infty}_{-i\infty}dz\frac{\mathcal{M}(z)}{z}\lambda^z$ in which "λ" is a number which approaches zero. Is the following way correct or not?

First we write it as $\int^{i\infty}_{-i\infty}dz\frac{\mathcal{M}(z)}{z}e^{z\log{\lambda}}$ where $\lambda$ is some meromorphic function, but on the exponential the first derivative of the exponent doesn't have any zero, therefore I pull the $1/z$ factor onto the exponent: $\int^{i\infty}_{-i\infty}dz\mathcal{M}(z)e^{z\log{\lambda}-\log{z}}$, then the exponent $z\log{\lambda}-\log{z}$ is stationary at z∼0 when λ→0, then we just approximate the integral with the limit of the integrand when z→0, which is $\mathcal{M}(0)\log{\lambda}$.

Is this way of doing steepest descent reasonable?

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user user106592
asked Nov 9, 2013 in Theoretical Physics by user106592 (35 points) [ no revision ]

1 Answer

+ 0 like - 0 dislike

This does not seem reasonable, at least not at first glance. It's hard to believe that the integral does not depend on the behavior of $M(z)$ near $z=0$. Maybe you should move $M(z)$ to the exponent as well.

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli
answered Nov 9, 2013 by akhmeteli (40 points) [ no revision ]
Yes, it is possible to include $\mathcal{M}(s)$ in the exponent, the result of the saddle point turns into $\log{\lambda}-\frac{1}{z}+\frac{\mathcal{M}'}{\mathcal{M}}=0$ and I assume $\frac{\mathcal{M}'}{\mathcal{M}}$ is not singular around zero. What I worry about is, the usual form of the exponent for steepest descent has an overall big coefficient like $\int dz A(z) e^{\lambda B(z)}$, but here the form of the exponential seems to be non-standard. I'm not sure if there are pitfalls.

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user user106592
And why do you assume that the fraction is not singular around zero? What if $M=z$?

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli
A better way of saying is, I'm doing the integral in this way for those $\mathcal{M}$ that satisfies such regularity condition, not for a general $\mathcal{M}$, in the problem I'm facing with, actually I don't know what $\mathcal{M}$ is, so I have to modestly impose some condtions first so that I can somehow proceed. So let's first take a regular $\frac{\mathcal{M}'}{\mathcal{M}}$ :)

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user user106592
OK, this assumption does not look reasonable to me, but you know better what you need...

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli
@user106592: You may wish to look at en.wikipedia.org/wiki/Stationary_phase_approximation

This post imported from StackExchange Physics at 2014-03-07 13:45 (UCT), posted by SE-user akhmeteli

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...