Q:

How to prove this equality for $\int_{\mathbb{R}_{+}^{n}}e^{-\left\langle a,x\right\rangle }dx$?

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?

A:

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.

