# Schwinger trick and following change of variables

1. Mar 20, 2012

### André H Gomes

1. The problem statement, all variables and given/known data

We have the evaluation of the integral associated with the electron self-energy diagram and I am following Brian Hatfield's book "Quantum Field Theory of point particles and strings", and I am having problems with the integration limits after changing variables from $z_2$ to $1-z_2$, at p.388.

2. Relevant equations

$\int_0^\infty \frac{d\beta}{\beta} \int_0^\infty dz_2 \left[2m_o - (1-z_2)\gamma^\mu p_\mu)\right] e^{-\beta[(1-z_2)z_2 p^2 - m^2_o z_2 - \lambda^2(1-z_2)]}$

Since it may be relevant, some additional information: before this integral, we had an integration over $z_1$, which involved a Dirac delta function, $\delta(1-z_1-z_2)$.

3. The attempt at a solution

I expected the change of variables to be as straightfoward as it seemed to be, resulting in:

$\int_0^\infty \frac{d\beta}{\beta} \int_{-\infty}^1 dz \left(2m_o - z\gamma^\mu p_\mu)\right) e^{-\beta[z(1-z) p^2 - m^2_o(1-z) - \lambda^2z]}$

But, in Hatfield's book, the lower limit of the integral in $z$ becomes $0$ instead of $-\infty$. I wonder if the integration from $-\infty$ to $0$ may vanish somehow, but I just can't figure it.

In advance, thank you all for the help!

(English is not my native language. I apologize any mistake.)

2. Mar 20, 2012

### André H Gomes

If think I found my mistake: evaluating the previous $z_1$ integration with the delta function will automatically change the integration limits of the $z_2$ integration.

Looking at the step before the above integral, where we have the integral over $z_1$, from $z_1=0$ to $z_1=\infty$, we note that:

- There is the delta, $\delta(1-z_1-z_2)$, demanding $z_1=1-z_2$.

- But $z_2$ is integrated from $z_2=0$ to $z_2=\infty$, saying that $z_1$ may assume values between $-\infty$ and $1$.

- On the other hand the integration over $z_1$ puts its values between $0$ and $\infty$.

- The range of values $z_2$ may assume so that the delta function argument is satisfied goes from $0$ to $1$.

Therefore, this should be the range of integration for the $z_2$ integration after the evaluation of the delta function: from $z_2=0$ to $z_2=1$. In such a way, the change of variables from $z_2$ to $z=1-z_2$ is indeed as straightfoward as it seems, and also the integration limits are as expected.