• 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.


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

202 submissions , 160 unreviewed
4,981 questions , 2,140 unanswered
5,340 answers , 22,629 comments
1,470 users with positive rep
813 active unimported users
More ...

  Integral manipulation and Lagrangian simplification

+ 1 like - 0 dislike

I'm trying to work out some expressions from [this paper](https://arxiv.org/abs/2008.02770), namely expression (4) from (11) and (10).

Consider the Lagrangian

  \mathcal{L}(M,W,\lambda) = g^2\left(1+\pi W(m^2)\right)+\int_{4m^2}^{\infty}ds\, \Im{\left[W(s)M(s)\right]}\tag{1}

Using the fact that 

 W(s) = \frac{1}{\pi}\int_{4m^2}^{\infty}dz \,\frac{\omega(z)}{s-z+i0}-\frac{\omega(z)}{4m^2-s-z+i0}\tag{2}

and the Sokhotski–Plemelj theorem,

\frac{1}{s-z\pm i0}= \mp i\pi \delta(s-z)+\mathcal{P}\frac{1}{(s-z)}\tag{3}

I am trying to prove that (1) is equivalent to

\mathcal{L}(M,\omega,\lambda) = g^2+\int_{m^2}^{\infty}ds\,\omega(s)\mathcal{A}(s)\tag{4}


\mathcal{A}(s) = M(s)-\left(M_{\infty}-\frac{g^2}{s-m^2}+\int_{4m^2}^{\infty}\frac{dz}{\pi}\, \frac{\Im{[M(z)]}}{s-z+i0}+\frac{\Im{[M(z)]}}{4-s-z+i0}\right)\tag{5}

**My attempt**

First, using the second to last equation in the Appendix B.1 of the paper,

W(z) = \frac{1}{\pi}\int_{4m^2}^{\infty}ds\, \Im{ [W(s)]}\left(\frac{1}{s-z}-\frac{1}{s-t(z)}\right)\tag{6}

where $t(z) = 4-z$ and $\Im{[W(s)]} = \omega(s)$ for $s > 4m^2$:

g^2\pi W(m^2) &= g^2\pi \times\frac{1}{\pi}\int_{4m^2}^{\infty}ds\,\omega(s)\left(\frac{1}{s-m^2}+\frac{1}{4-s-m^2}\right)\\
&= -\frac{1}{\pi}\int_{4m^2}^{\infty}ds\,\omega(s)\left(-\frac{g^2}{s-m^2}-\frac{g^2}{4-s-m^2}\right)

For the integral, using $\Im{[M(s)\times W(s)]} = \Im{[M(s)]}\Re{[W(s)]}+\Re{[M(s)]}\Im{[W(s)]}$

\int_{4m^2}^{\infty}ds\, \Im{\left[W(s)M(s)\right]} = \int_{4m^2}^{\infty}ds\,\Im{[M(s)]}\left(W(s)-2i\Im{[W(s)]}\right)+\,\int_{4m^2}^{\infty}ds \Im{[W(s)]}M(s)

The second integral can be identified with 

\int_{4m^2}^{\infty}ds \Im{[W(s)]}M(s) = \int_{4m^2}^{\infty}ds \, \omega(s)M(s)

On the other hand, 

\int_{4m^2}^{\infty}ds\,\Im{[M(s)]}W(s) = \int_{4m^2}^{\infty}\omega(s)\int_{4m^2}^{\infty}\frac{dz}{\pi}\, \frac{\Im{[M(z)]}}{s-z+i0}-\frac{\Im{[M(z)]}}{4-s-z+i0}

where eq. (2) and a Dirac-delta were used.


 - I tried using the aforementioned theorem to simplify the remaining of the expression, but to no avail. A Cauchy principal value appears which I am not being able to deal with;
 - How is it that eqs. (2) and (6) match? I tried to simplify one of them, but I could not show that they were the same expression.

asked Feb 6, 2022 in Theoretical Physics by anonymous [ 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).
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