# Calculation of g-factor correction in Peskin p. 196

## Homework Statement:

calculation of the correction for g-factor

## Relevant Equations:

F_2(0)=(g-2)/2
I'm reading peskin QFT textbook. In page 196 eq. (6.58) it says

$$F_2(q^2=0)=\frac{\alpha}{2\pi}\int ^1_0 dx dy dz \delta (x+y+z-1) \frac{2m^2z(1-z)}{m^2(1-z)^2}\\=\frac{\alpha}{\pi}\int ^1_0 dz\int ^{1-z}_0 dy \frac{z}{1-z}=\frac{\alpha}{2\pi}$$

I confirmed the conversion from the first line to the second but can't figure out how to convert the second to the third. I think the last line should be α/π.
This is used to calculate the correction to g-factor

$$a_e=\frac{g-2}{2}=F_2(0)=\frac{\alpha}{2\pi}=0.0011614...$$

Is there something I'm missing or is it an error in the textbook?

Last edited:

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TSny
Homework Helper
Gold Member
I think the book is correct. After doing the ##y## integration, what does the ##z##-integral look like?

I think the book is correct. After doing the ##y## integration, what does the ##z##-integral look like?
From the second line,

$$\frac{\alpha}{\pi}\int ^1_0 dz \int^{1-z}_0 dy \frac{z}{1-z}=\frac{\alpha}{\pi}\int ^1_0 dz z=\frac{\alpha}{\pi}$$

TSny
Homework Helper
Gold Member
I agree with the form of your ##z## integral. But it looks like you're not doing the ##z##-integration correctly.

niss
I agree with the form of your ##z## integral. But it looks like you're not doing the ##z##-integration correctly.
What do you mean by "correctly"?

MathematicalPhysicist
Gold Member
What do you mean by "correctly"?
##\int zdz = z^2/2##.

niss
##\int zdz = z^2/2##.
OMG I seriously missed it. Sorry for posting such a stpd question...