Denote \begin{align*} & \mathbb{R}_{+}^{n}=\left\{ \left(x_{1},\ldots,x_{n}\right)\in\mathbb{R}^{n}\mid x_{i}>0\;,\;1\leq i\leq n\right\} \\ & S^{n-1}=\left\{ x\in\mathbb{R}^{n}\mid\left|x\right|=1\right\} \\ & S_{+}^{n-1}=S^{n-1}\cap\mathbb{R}_{+}^{n} \end{align*}
Im looking to prove that $$ \int_{\mathbb{R}_{+}^{n}}e^{-\left\langle a,x\right\rangle }dx_{1}\ldots dx_{n}=\left(n-1\right)!\int_{S_{+}^{n-1}}\frac{1}{\left\langle a,y\right\rangle ^{n}}dS\left(y\right) $$
Attempt:
Looking first on the RHS we see that \begin{align*} \left(n-1\right)!\int_{S_{+}^{n-1}}\frac{1}{\left\langle a,y\right\rangle ^{n}}dS\left(y\right) & =\Gamma\left(n\right)\int_{S_{+}^{n-1}}\frac{1}{\left\langle a,y\right\rangle ^{n}}dS\left(y\right)=\\ & =\int_{0}^{\infty}\rho^{n-1}e^{-\rho}d\rho\int_{S_{+}^{n-1}}\frac{1}{\left\langle a,y\right\rangle ^{n}}dS\left(y\right)=\\ & =\lim_{r\to\infty}\int_{0}^{r}\rho^{n-1}\left(\int_{S_{+}^{n-1}}\frac{e^{-\rho}}{\left\langle a,y\right\rangle ^{n}}dS\left(y\right)\right)d\rho \end{align*} Now as for the LHS we can take the exhaustion $\mathbb{R}_{+}^{n}=\bigcup_{r=1}^{\infty}rB_{+}^{n}$ where $B_{+}^{n}$ is the positive side of the unit ball, and using the Coarea formula we get \begin{align*} \int_{\mathbb{R}_{+}^{n}}e^{-\left\langle a,x\right\rangle }dx_{1}\ldots dx_{n} & =\lim_{r\to\infty}\int_{rB_{+}^{n}}e^{-\left\langle a,x\right\rangle }dx_{1}\ldots dx_{n}=\\ & =\lim_{r\to\infty}\int_{0}^{r}\left(\int_{\rho S_{+}^{n-1}}e^{-\left\langle a,x\right\rangle }dS\left(x\right)\right)d\rho=\\ & =\lim_{r\to\infty}\int_{0}^{r}\rho^{n-1}\left(\int_{S_{+}^{n-1}}e^{-\left\langle a,\rho y\right\rangle }dS\left(y\right)\right)d\rho=\\ & =\lim_{r\to\infty}\int_{0}^{r}\rho^{n-1}\left(\int_{S_{+}^{n-1}}e^{-\rho\left\langle a,y\right\rangle }dS\left(y\right)\right)d\rho \end{align*} Is this right? How can I move on from here to get the equality?
You're almost there. A scaling gives $$\int_0^r X^{n-1}\int_{S_+^{n-1}}\frac{e^{-X}}{\langle a,\,y\rangle^n}dS dX=\int_0^{r/\langle a,\,y\rangle}Y^{n-1}\int_{S_+^{n-1}}e^{-Y\langle a,\,y\rangle}dS dY,$$and in imposing $r\to\infty$ we can change the upper limit $r/\langle a,\,y\rangle$ to $r$, then use $\rho$ as a replacement label for both $X$ and $Y$. (We just require that $\langle a,\,y\rangle>0$, which is true if each component of $a$ is positive. If you write $\int_0^\infty d\rho$ instead of $\lim_{r\to\infty}\int_0^r d\rho$, it's a little more obvious.) This equates the limits you need to equate.