Calculation of g-factor correction in Peskin p. 196

[niss]

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?

 

[TSny]

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

 

[niss]

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]

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]

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...