Feynman Parameters--Peskin&Schroeder 6.44 I'm having some trouble deriving equation 6.44 on p.191 of Peskin and Schroeder (my book is the fifth printing, December 1997). The algebra comes close but I think that you need to do some arguing to actually get his answer--maybe this requires some Dirac equation reasoning that I'm not seeing. Here's the problem: We're working on a Feynman integral for electrons and photons, so everything is sandwiched between u(p,s) electron state-functions. The precise step I'm on is converting the denominator of the integral into the integral ∫dx dy dz δ(1-x-y-z) 2/D3. We are given that: D = k2+2k(yq-zp)+yq2+zp2-(x+y)m2+iε With the Dirac delta in the integral, we can always assume 1=x+y+z. He makes the definition l=k+yq-zp. Then he says So I tried doing the algebra, and immediately it seems we need to use the Dirac equation to get another m2 term. Starting with the given D, I got D=l2-[z2p2]+2zpyq-y2q2+yq2+[zp2]-(1-z)m2+iε So after staring at that a while, I decided there's no way to get anything else into the coefficient of m2 without applying the Dirac equation. So I made the substitution p2 = m2 in the bracketed terms. [This is the only time I applied the Dirac equation. Is it valid?] This gets us closer--it changes the coefficient of m2 into what we want. A few steps later I found: D=l2+2zpyq+yzq2+yxq2-(1-z)2m2+iε =l2-Δ+iε+yzq2+2zpyq =l2-Δ+iε+yz[(p+q)2-p2] Aside from that one application of the Dirac equation, I didn't use anything but algebra. Why are those pesky terms there? Should we be saying they cancel, somehow, or that q=0? (In Ryder's QFT, he mentions something like this on p.344, but I don't really get it.) If that's the case, why would q not drop out of the Δ term? Did I misapply the Dirac equation? Thanks for any and all help!