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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?

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    $\begingroup$ +1 for all your work. As a side note, you should use $\prod$ (\prod) instead of $\Pi$ (\Pi). $\endgroup$ – JMCF125 Jan 19 '14 at 21:12
  • $\begingroup$ 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. $\endgroup$ – Stephen Montgomery-Smith Apr 12 '14 at 4:21
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$$ \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n} $$ $$ = \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n} \int_{s=0}^\infty ds \delta\left(\sum_{j=1}^n y_j - s\right) $$ $$ = \int_{s=0}^\infty ds \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n}\delta\left(\sum_{j=1}^n y_j - s\right) $$ 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 $$ \int_0^\infty dy_1 \dots \int_0^\infty dy_n e^{-A_1 y_1-\cdots-A_n y_n} $$ $$ = \prod_{j=1}^n \int_0^\infty dy_j e^{-A_j y_j} = \left(\prod_{j=1}^n A_j \right)^{-1} .$$

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  • $\begingroup$ I do not understand those first two steps in the slightest. $\endgroup$ – Señor O Jan 18 '18 at 3:41
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    $\begingroup$ I was missing a bunch of exponentials in the second step. I put them in. Sorry. $\endgroup$ – Stephen Montgomery-Smith Jan 18 '18 at 3:46
  • $\begingroup$ 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. $\endgroup$ – Stephen Montgomery-Smith Jan 18 '18 at 3:47
  • $\begingroup$ Thank you, I appreciate the update and the proof. +1 $\endgroup$ – Señor O Jan 18 '18 at 6:04

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