# Feynman Parameters

I'm trying to prove the following identity: $$\left(\prod_{j=1}^n A_j\right)^{-1} = \int_0^1dx_1 \dots \int_0^1dx_n \,\delta\left(\sum_{i=1}^{n}x_i -1\right) \frac{(n-1)!}{(\sum_{j=1}^nx_iA_i)^{n}}\,\,\,\,\,\,\,\,\forall n\in\mathbb{N}$$ My strategy is induction on $n$. The $n=1$ step is easy to show. Assuming the induction hypothesis, we have: $$\left(\prod_{j=1}^{n+1} A_j\right)^{-1} = \left(\Pi_{j=1}^{n} A_j\right)^{-1} \cdot \left( A_{n+1} \right)^{-1} = \int_0^1dx_1 \dots \int_0^1dx_n \,\delta\left(\sum_{i=1}^{n}x_i -1\right)\left((n-1)!\right)\frac{1}{A_{n+1}(\sum_{j=1}^nx_iA_i)^{n}}$$ Then I use an identity which is easy to prove by differentiation of both sides of the equation repeatedly w.r.t. $B$: $$\frac{1}{A\cdot B^n} = \int_0^1dx\int_0^1dy\delta\left(x+y-1\right)\frac{n\,y^{n-1}}{\left(xA+yB\right)^{n+1}}\,\,\,\forall n\in\mathbb{N}$$ to obtain: $$\left(\prod_{j=1}^{n+1} A_j\right)^{-1} = \int_0^1dx_1 \dots \int_0^1dx_n \,\delta\left(\sum_{i=1}^{n}x_i -1\right)\left((n-1)!\right)\int_0^1dx\int_0^1dy\delta\left(x+y-1\right)\frac{ny^{n-1}}{\left(xA_{n+1}+y\sum_{i=1}^nx_iA_i\right)^{n+1}}$$ Perform the $x$ integration to get: $$\left(\prod_{j=1}^{n+1} A_j\right)^{-1} = \int_0^1dx_1 \dots \int_0^1dx_n \,\delta\left(\sum_{i=1}^{n}x_i -1\right)\left((n-1)!\right)\int_0^1dy\frac{ny^{n-1}}{\left(\left(1-y\right)A_{n+1}+y\sum_{i=1}^nx_iA_i\right)^{n+1}}\\\ = \int_0^1dy\int_0^1dx_1 \dots \int_0^1dx_n \,\delta\left(\sum_{i=1}^{n}x_i -1\right)\left((n-1)!\right)\frac{ny^{n-1}}{\left(\left(1-y\right)A_{n+1}+y\sum_{i=1}^nx_iA_i\right)^{n+1}}$$ Now make the following $n$ substitutions: $$u_i := y x_i \forall i \in \left\{1,\dots,n \right\}$$ Which results in: $$= \int_0^1dy\int_0^y\frac{du_1}{y} \dots \int_0^y\frac{du_n}{y} \,\delta\left(\sum_{i=1}^{n}\frac{u_i}{y} -1\right)\left((n-1)!\right)\frac{ny^{n-1}}{\left(\left(1-y\right)A_{n+1}+\sum_{i=1}^n u_iA_i\right)^{n+1}}$$ Use the delta function identity: $$\delta\left(\sum_{i=1}^n \frac{u_i}{y} -1\right) = \delta\left(\sum_{i=1}^n u_i -y\right)\cdot y$$ To get: $$= \int_0^1dy\int_0^y du_1 \dots \int_0^y du_n \,\delta\left(\sum_{i=1}^{n}u_i -y\right)\frac{n!}{\left(\left(1-y\right)A_{n+1}+\sum_{i=1}^n u_iA_i\right)^{n+1}}$$ Finally, make the substitution $u_{n+1} := 1-y$ $$= \int_0^1du_{n+1}\int_0^{1-u_{n+1}} du_1 \dots \int_0^{1-u_{n+1}} du_n \,\delta\left(\sum_{i=1}^{n+1}u_i -1\right)\frac{n!}{\left(\sum_{i=1}^{n+1} u_iA_i\right)^{n+1}}$$ Now I am stuck, because I don't know how to deal with the integration limits. I could stretch them all from $0$ to $1$, but as far as I can tell, that would mean I'd have to divide the whole thing by $n!$, which would not give me the result I'm looking for.

What am I doing wrong?

• +1 for all your work. As a side note, you should use $\prod$ (\prod) instead of $\Pi$ (\Pi). – JMCF125 Jan 19 '14 at 21:12
• In your last displayed equation, you can see that when you change each of the inner integrals from $\int_0^{1-u_{n+1}}$ to $\int_0^1$ (not by stretching, but just by integrating over the larger interval), that the extra part you are integrating over is zero. – Stephen Montgomery-Smith Apr 12 '14 at 4:21

$$\int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n}$$ $$= \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n} \int_{s=0}^\infty ds \delta\left(\sum_{j=1}^n y_j - s\right)$$ $$= \int_{s=0}^\infty ds \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n}\delta\left(\sum_{j=1}^n y_j - s\right)$$ Make the substitution $y_j = s x_j$, and note $\delta(sx) = s^{-1} \delta(x)$ if $s > 0$, to get $$= \int_{s=0}^\infty ds \, s^{n-1} \int_0^\infty dx_1 \dots \int_0^\infty dx_n e^{-(A_1 x_1-\cdots-A_n x_n)s}\delta\left(\sum_{j=1}^n x_j - 1\right)$$ $$= \int_0^\infty dx_1 \dots \int_0^\infty dx_n \delta\left(\sum_{j=1}^n x_j - 1\right) \int_{s=0}^\infty ds \, s^{n-1} e^{-(A_1 x_1-\cdots-A_n x_n)s}$$ and evaluate the integral over $s$ to get $$= \int_0^\infty dx_1 \dots \int_0^\infty dx_n \delta\left(\sum_{j=1}^n x_j - 1\right) \frac{(n-1)!}{(\sum_{j=1}^nx_jA_j)^{n}}$$ But also $$\int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n}$$ $$= \prod_{j=1}^n \int_0^\infty dy_j e^{-A_j y_j} = \left(\prod_{j=1}^n A_j \right)^{-1} .$$
• The second step simply uses $\int_{s=0}^\infty ds \delta\left(\sum_{j=1}^n y_j - s\right)=1$. The third step is rearranging the integrals. – Stephen Montgomery-Smith Jan 18 '18 at 3:47