# Normalization of the integration measure of the Feynman's formula to combine denominators

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?

• To prove (14.11) introduce new variables, y1=x1, y2=x2, ... and yn= x1+x2+ ... + xn. – Oбжорoв Jun 7 at 8:09
• @Oбжорoв. Can you explicit a little more? – Michele Grosso Jun 7 at 14:42
• Change variables in the integration. If it is not clear, first work out a few cases, n=1,2,3. – Oбжорoв Jun 9 at 9:46

## 1 Answer

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.

• 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? – Michele Grosso Jun 7 at 14:41
• 14.11 follows immediately: set all the $A_i$ to unity, then $\sum_i A_i x_i=1$ since the delta function enforces $\sum_i x_i=1$. – mike stone Jun 7 at 20:52
• I did not realize the last step. Thanks! – Michele Grosso Jun 8 at 7:20