Integral Analysis: Equality of Lorentz Structures

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SUMMARY

The integral discussed, \int d^{3}p_{2} \frac{1}{2E_{2}} d^{3}p_{3} \frac{1}{2E_{3}} \delta^{4}(p-p_{2}-p_{3}) p_{3}^{\alpha} p_{3}^{\beta} p_{2}^{\mu}, is confirmed to be equal to the expression A(p^{2})g^{\alpha\beta}p^{\mu}+B(p^{2})g^{\beta\mu}p^{\alpha}+B(p^{2})g^{\mu\alpha}p^{\beta}, where the coefficients of the last two terms are identical due to the symmetry of the integral variables p_{3}^{\alpha} p_{3}^{\beta}. The discussion emphasizes the importance of using LaTeX formatting for clarity in communication, which can enhance readability and response time in forums.

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Hi, I am a newcomer, and i have a question about an integral shown as follows,

[tex]\int d^{3}p_{2} \frac{1}{2E_{2}} d^{3}p_{3} \frac{1}{2E_{3}} \delta^{4}(p-p_{2}-p_{3}) p_{3}^{\alpha} p_{3}^{\beta} p_{2}^{\mu}[/tex]

Is this integral equal to (just considering Lorentz structures)

[tex]A(p^{2})g^{\alpha\beta}p^{\mu}+B(p^{2})g^{\beta\mu}p^{\alpha}+B(p^{2})g^{\mu\alpha}p^{\beta}[/tex]

where the coefficients of the last two terms should be the same because of the symmetry of the integral variables [tex]p_{3}^{\alpha} p_{3}^{\beta}[/tex]

Thank u for ur help!
 
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It's been a while since I did such HEP integrals, so I can't give you an answer off the top of my head. But I want to give you a tip, since you have posted the formulas in valid LaTeX code anyway: if you wrap them in tex-tags (put [ tex] at the beginning and [/ tex] at the end, and take out the spaces in both :)) they will be nicely rendered by LaTeX. For inline (i.e. $ ... $) you can replace tex by itex in both tags. That will make your post a lot more readable, probably giving you faster responses as well.

Quote my message to see the codes used:
[tex]\int d^{3}p_{2} \frac{1}{2E_{2}} d^{3}p_{3} \frac{1}{2E_{3}} \delta^{4}(p-p_{2}-p_{3}) p_{3}^{\alpha} p_{3}^{\beta} p_{2}^{\mu}[/tex]
[...] integral variables [itex]p_{3}^{\alpha} p_{3}^{\beta}[/itex].
 

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