Error in Standard Textbook "An Introduction to Quantum Field Theory" of Peskin and Schroeder? On page 191 there is a equation for $D$ given by
$$D=x(k^2-m^2)+y(k'^2-m^2)+z(k-p)^2+(x+y+z)i\epsilon. \tag{6.43}$$
With $k'=k+q$ and the constraint $x+y+z=1$.
Also $p^2=p'^2=m^2$ and maybe $q=p'-p$.
Now he completes the square and shifts $k$ while he introduces $$\ell\equiv k+yq-zp\qquad$$ 
(Substituting $\ell$ for $k$.)
Then you can write $D$ as $$D=\ell^2-\Delta+i\epsilon.$$
My question is: What do you get for $\Delta$?
My friends and I calculated this some times and never got the result given in the text book. (The differences does not influence the result at the very end..)
 A: Hint: Recall that
$$\tag{A} q\cdot (q+2p) ~=~(p^{\prime}-p)\cdot (p^{\prime}+p)
~=~p^{\prime 2}-p^2~=~m^2-m^2~=~0 .$$
So 
$$\tag{B}  \Delta -(1-z)^2m^2~=~y(y-1)q^2 -2yz q\cdot p~\stackrel{(A)}{=}~y(y-1+z)q^2~=~-xyq^2. $$
A: When I did the derivation I got this for delta:
$\Delta = -y^2q^2 +yq^2 -2zpyq +(1-z)^2m^2$
In the text it says this:
$\Delta = -xyq^2+(1-z)^2m^2$
Is there some information missing?
I recommend looking up "how to use Feynman Parameters" on google to get more detail, but basically it looks like the derivation is given on this webpage, in example 2.  It looks like it was checked in Maple or Mathematica.
http://theoretical-physics.net/dev/src/math/feynman-parameters.html
A: I remember being confused for a while as well when I got to that part. Below is my step-by-step calculation. Although this post is a bit old, I think a non-zero amount of people also have this question and my answer will be useful.
$$\begin{equation}
D=k^{2}+2k\cdot(yq-zp)+yq^{2}+zp^{2}-(x+y)m^{2}+i\epsilon
\end{equation}$$
We want to complete the square, so define $l=k+(yq-zp)$.
$$\begin{equation}
D=l^{2}-(yq-zp)^{2}+yq^{2}+zp^{2}-(x+y)m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}-y^{2}q^{2}+2yzq\cdot p -z^{2}p^{2}+yq^{2}+zp^{2}-(x+y)m^{2}+i\epsilon
\end{equation}$$
Now, look back at the diagram at the start of Section 6.3. Recall that $p$ and $p'$ correspond to external legs. This means that $p^{2}=m^{2}$ and $p'^{2}=m^{2}$. Also, $x+y+z=1$ in the parametrization.
Plug in $p^{2}=m^{2}$ and use $x+y+z=1$.
$$\begin{equation}
D=l^{2}-y^{2}q^{2}+2yzq\cdot p -z^{2}m^{2}+yq^{2}+zm^{2}-(x+y)m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}-y^{2}q^{2}+2yzq\cdot p +yq^{2}-(x+y-z+z^{2})m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}-y^{2}q^{2}+2yzq\cdot p +yq^{2}-(x+y+z-2z+z^{2})m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}-y^{2}q^{2}+2yzq\cdot p +yq^{2}-(1-2z+z^{2})m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}-y^{2}q^{2}+2yzq\cdot p +yq^{2}-(1-z)^{2}m^{2}+i\epsilon
\end{equation}$$
This is almost what the textbook has. The other fact that we need to use is that $p'^{2}=m^{2}$ Look at the diagram at the start of Section 6.3 and convince yourself that $p'=p+q$. This means that $(p+q)^{2}=m^{2}$, so that $p^{2}+2p\cdot q +q^{2}=m^{2}$, and hence $2p\cdot q=-q^{2}$. We can use this to rewrite the $q\cdot p$ term in the expression for $D$.
$$\begin{equation}
D=l^{2}-y^{2}q^{2}-yzq^{2} +yq^{2}-(1-z)^{2}m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}+(-y^{2}-yz+y)q^{2} -(1-z)^{2}m^{2}+i\epsilon
\end{equation}$$
$$\begin{equation}
D=l^{2}+y(-y-z+1)q^{2}-(1-z)^{2}m^{2}+i\epsilon
\end{equation}$$
Use $x+y+z=1$ again.
$$\begin{equation}
D=l^{2}+yxq^{2} -(1-z)^{2}m^{2}+i\epsilon
\end{equation}$$
If you write $D\equiv l^{2}-\Delta+i\epsilon$, we can easily read off that $\Delta \equiv -yxq^{2}+(1-z)^{2}m^{2}$. This is what the textbook has.
