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Look at

[tex]

\begin{align}

& \int d^4 x d^4 x_1 d^4 x_2 d^4 x_3 d^4 x_4 \exp [i(p_1 x_1 + p_2 x_2 -k_1 x_3 -k_2 x_4)] \\

& \times (-i\lambda)D(x_1-x)D(x_2-x)D(x_3-x)D(x_4-x)

\end{align}

[/tex]

for first order in lambda for 2-2 scattering. In Maggiore I am told to substitute [itex]y_i=x_i-x[/itex] as a variable substitution and then carry out the [itex]y_i[/itex] integrals. Fair enough, but what about the measures? [itex]y_i=x_i -x \rightarrow dy_i = dx_i -dx [/itex]. Surely [itex]x[/itex] isn't a constant, since the coupling point could be anywhere, then I should have a whole separate integral with [itex]-d^4 x[/itex], right? If I make the substitution while ignoring the measure for emphesis I get

[tex]

\begin{align}

& \int d^4 x d^4 x_1 d^4 x_2 d^4 x_3 d^4 x_4 \exp [i(p_1 (y_1+x) + p_2 (y_2+x) -k_1 (y_3+x) -k_2 (y_4+x))] \\

& \times (-i\lambda)D(y_1)D(y_2)D(y_3)D(y_4)

\end{align}

[/tex]

now if I directly make the substitution [itex]d^4 x_i \rightarrow d^4 y_i [/itex] and integrate over [itex]y_i[/itex] I get

[tex]

(-i\lambda) D(p_1)D(p_2)D(p_3)D(p_4) \int d^4 x \exp[i(p_1 + p_2 - k_1 -k_2)x]

[/tex]

which is precisely the next line of Maggiore, except I cheated, didn't I...? (Here [itex]D(p_i)[/itex] are the momentum space representation of the Feynman propagator.) What happened to my [itex]-d^4 x[/itex] when I make the variable substitution above? A point in the right direction would be lovely, thanks.

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# 4-point correlation in phi 4 to first order

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