• 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.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,734 comments
1,470 users with positive rep
818 active unimported users
More ...

  Dimensional Regularization Integral Formula

+ 5 like - 0 dislike

In the formula $$\int \frac {d^{4-2\epsilon} l} {(2\pi)^{4-2\epsilon}} \frac 1 {(l^2-\Delta)^2} = \frac i {(4\pi)^{2-\epsilon}} \Gamma(\epsilon) \left(\frac 1 \Delta\right)^\epsilon,$$ how should I deal with the case when $\Delta<0$?

This post has been migrated from (A51.SE)
asked Dec 17, 2011 in Theoretical Physics by felix (110 points) [ no revision ]

2 Answers

+ 2 like - 0 dislike

In complex calculus, one may compute powers – and other functions – for all negative or complex values of the argument. But it would lead to confusions and ambiguities and indeed, it's not needed.

Whenever the theory is stable, it's guaranteed that $\Delta$ in these d.r. integrals ends up positive. A typical value of $\Delta$ for Feynman-parameterized integrals is an example from a 4-point diagram (imagine 2 to 2 scattering) $$ \Delta = m^2 - x(1-x)q^2 $$ Now, $m^2$ is positive because we don't have tachyons and $x(1-x)$ is between $0$ and $1/4$. The only risk how the second term could beat $m^2$ and make $\Delta$ negative is that $q$ is hugely timelike, with length over $2m$. But it really can't happen because $q$ is the sum of two external momenta. Timelike $2m$ length is really the maximum and in most cases, $q$ will end up shorter or spacelike. The UV region we're integrating corresponds to very long spacelike $q$ (all directions in the Euclideanized spacetime are spacelike) for which $\Delta$ is high and positive because the spacelike vectors have negative signature above.

I can't provide you with a general proof (it may be easy, however) but I am pretty sure that the negative-$\Delta$ problem doesn't arise in any loop diagrams.

This post has been migrated from (A51.SE)
answered Dec 17, 2011 by Luboš Motl (10,278 points) [ no revision ]
I have an example where $\Delta$ is negative. Consider the one-loop EM vertex correction for fermion-fermion scattering in QED. Ignoring fermion masses, we have $\Delta=-x(1-x)q^2$, where $q^2$ is the photon virtuality. Now $q^2$ is -ve for t-channel elastic scattering, but +ve for s-channel scattering, so $\Delta$ may have either signs.

This post has been migrated from (A51.SE)
+ 2 like - 0 dislike

I point out that I agree with the preceding answer but the question here is just a mathematical one. The value of the integral will depend on the sign of $\Delta$. So, if $\Delta<0$ the integral value is not the one you wrote but

$$\int dx\frac{x^{3-2\epsilon}}{(x+\Delta)^2}=x^{2-2\epsilon}\left(\frac{1}{x-\Delta^2}\right)^{2\epsilon-2}\ _2F_1(2\epsilon-2, 2\epsilon-3, 2\epsilon-1, -\Delta^2/(x - \Delta^2))\times$$ $$\frac{1}{2-2\epsilon}$$

The beast that you see on the rhs is a hypergeometric function that does not converge when you take the limit of $x$ going to infinity. So, physical arguments given above are sound.

This post has been migrated from (A51.SE)
answered Dec 17, 2011 by JonLester (345 points) [ no revision ]

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:
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).
Please complete the anti-spam verification

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

Your rights