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In Mark Srednicki "Quantum field theory", section 14 -Loop corrections to the propagator-, it is presented the Feynman's formula to combine denominators:
$\frac{1}{A_1 ... A_n} = \int dF_n (x_1 A_1 + ... + x_n A_n)^{-n}$ Eq. (14.9)
where the integration measure $dF_n$ over the Feynman's parameters $x_i$ is
$\int dF_n = (n - 1)! \int_0 ^1 dx_1 ... dx_n \delta (x_1 + ... + x_n - 1)$ Eq. (14.10)
The measure is normalized so that
$\int dF_n 1 = 1$ Eq. (14.11)

While Eq. (14.9) is given a hint to prove it in problem 14.1 (I succeeded to prove it), no hint is given to Eq. (14.11), that is the normalization of the integration measure $dF_n$.

My question is: How to prove Eq. (14.11), that is the normalization of the integration measure $dF_n$? Or, is there any reference (link) where I can find the demonstration?

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  • $\begingroup$ To prove (14.11) introduce new variables, y1=x1, y2=x2, ... and yn= x1+x2+ ... + xn. $\endgroup$ – Oбжорoв Jun 7 at 8:09
  • $\begingroup$ @Oбжорoв. Can you explicit a little more? $\endgroup$ – Michele Grosso Jun 7 at 14:42
  • $\begingroup$ Change variables in the integration. If it is not clear, first work out a few cases, n=1,2,3. $\endgroup$ – Oбжорoв Jun 9 at 9:46
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Start from Schwinger's trick $$ \frac{1}{A_1\ldots A_n}= \int_0^\infty dt_1\cdots dt_n e^{-\sum_{i=1}^\infty t_i A_i} $$ and insert $$ 1= \int_0^\infty d\tau \,\delta(t_1+\ldots+ t_n-\tau) $$ into the integral. (For each set of $t_i$ in the integration domain there is always has a unique value of $\tau$ for which $t_1+\ldots+ t_n=\tau$) to get $$ \frac{1}{A_1\ldots A_n}= \int_0^\infty d\tau \int_0^\infty d^n t \delta(t_1+\ldots+ t_n-\tau) e^{-\sum_i t_i A_i} $$ now write $t_i = \tau x_i$ and use $\delta(x\tau)=\tau^{-1}\delta(x)$ in the form $$ \delta(\tau (x_1+\ldots+ x_n -1))= \tau^{-1} \delta(x_1+\ldots x_n -1) $$ to get $$ \frac{1}{A_1\ldots A_n}=\int_0^\infty \tau^{n-1} d\tau\left\{ \int_0^\infty d^n x \,\delta(x_1+\ldots x_n -1)e^{-\tau(\sum_i x_i A_i)}\right\}\\ = \Gamma(n) \int_0^\infty d^nx \delta(x_1+\ldots x_n -1)\frac 1 {(\sum x_i A_i)^{n}}. $$ Then, since $\Gamma(n)=(n-1)!$, Feynman's result follows.

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  • $\begingroup$ What you proved is Eq. (14.9), which I already succeeded to prove, and not Eq. (14.11) as I asked for. Can you prove what I asked? $\endgroup$ – Michele Grosso Jun 7 at 14:41
  • $\begingroup$ 14.11 follows immediately: set all the $A_i$ to unity, then $\sum_i A_i x_i=1$ since the delta function enforces $\sum_i x_i=1$. $\endgroup$ – mike stone Jun 7 at 20:52
  • $\begingroup$ I did not realize the last step. Thanks! $\endgroup$ – Michele Grosso Jun 8 at 7:20

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