Show that $\int_{S^{n-1}_+}\frac{1}{\left \langle y,a \right \rangle^n}dS(y)=\frac{1}{(n-1)!}\frac{1}{a_1\dots a_n}$

Question

Let $\mathbb{R}^n_{+}$ be the set of points in $\mathbb{R}^n$ with all coordinates positive.
Let $S^{n-1}_{+}=S^{n-1}\cap \mathbb{R}^n_{+}$. Let $a=(a_1, \dots ,a_n)\in \mathbb{R}^n_{+}$. Show that: $$\int_{S^{n-1}_+}\frac{1}{\left \langle y,a \right \rangle^n}dS(y)=\frac{1}{(n-1)!}\frac{1}{a_1\dots a_n}$$ Hint: Compute the integral $$\int_{\mathbb{R}^n_+}e^{-\left \langle x,a \right \rangle}dx_1\dots dx_n$$ in two different ways.

I've tried to use the hint. The easy way to compute the integral is by Fubini: $$\int_{\mathbb{R}^n_+}e^{-\left \langle x,a \right \rangle}dx_1\dots dx_n= \left ( \int_{0}^{\infty}e^{-a_1x_1}dx_1 \right )\dots \left ( \int_{0}^{\infty}e^{-a_nx_n}dx_n \right )=\frac{1}{a_1}\dots\frac{1}{a_n}$$ This is a question under the title of "co-area fomula" so the second way should be using the formula to switch to integration on the positive side of all spheres centered at the origin.
But that gives: $$\int_{\mathbb{R}^n_+}e^{-\left \langle x,a \right \rangle}dx_1\dots dx_n= \int_0^{\infty}\left ( \int_{\rho S_{+}^{n-1}}e^{-\left \langle x,a \right \rangle}dS(x)\right ) d\rho$$ And I can't see how on earth can such an expression lead to the desired equality.

No one I know has managed to solve this. I'd very appriciate your help.
Good solutions that do not use the hint will also be welcomed.

Answer 1

Let $I$ denote your last integral and notice that, under the scaling $x \mapsto \rho x$, it can be written as

$$ I = \int_{0}^{\infty} \int_{S_+^{n-1}} e^{-\rho \langle x, a\rangle} \rho^{n-1} \,dS(x) d\rho = \int_{S_+^{n-1}} \left( \int_{0}^{\infty} \rho^{n-1} e^{-\langle x, a\rangle \rho} \, d\rho \right) \, dS(x) $$

because now two integral is decoupled.^[1] Now the inner integral is just the Laplace transform of the function $\rho \mapsto \rho^{n-1}$ evaluated at $\langle x, a\rangle$,^[2] hence yielding

$$ I = \int_{S_+^{n-1}} \frac{(n-1)!}{\langle x, a\rangle^n} \, dS(x) $$

as desired.

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Footnotes.

1. The change of order of integration is justified because the integrand is non-negative. This follows from Tonelli's theorem, which tells that if the integrand is non-negative then you can interchange the order of two integrals unconditionally.

2. If you are not familiar to this integral, just consider the sequence

   $$ I_n = \int_{0}^{\infty} x^n e^{-sx} \, dx $$

   for a fixed $s > 0$ and use integration by parts to show that the following recursive formula is satisfied:

   $$ I_0 = \frac{1}{s}, \qquad I_{n} = \frac{n}{s} I_{n-1} \quad \forall n \geq 1. $$

   Therefore $I_n = \frac{n!}{s^{n+1}}$ and our case corresponds to $I_{n-1}$ with $s = \langle x, a\rangle$.