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Denote $\Bbb R^n_+:=\{(x_1,...,x_n): x_i > 0 \;\;\; 1\leq i \leq n \}$. Prove:

$$\int_{\Bbb R^n_+}e^{-\langle a,x \rangle}dx_1...dx_n = \frac{1}{a_1\cdot... \cdot a_n}$$


What I tried:

$$\int_{\Bbb R^n_+}e^{-\langle a,x \rangle}dx_1...dx_n =\lim_{t \to \infty}\int_{(0,t)^n}e^{-\langle a,x \rangle}dx_1...dx_n = \prod_{i=1}^n \int_0^\infty e^{-a_ix_i}dx_i$$

From here it's pretty easy to finish, but the problem I'm facing is explaining why it is ok to assume that $a_i > 0$ for all $i$. Is it even true? or do I need to be told that in the question?

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    $\begingroup$ The assumption that $a_i>0$ for all $i$ cannot be missed, I think. $\endgroup$ – RMWGNE96 May 13 at 14:21
  • $\begingroup$ Maybe you can parametrize $\mathbb{R}_+^n$ by a radius $r>0$, $n-2$ angles $\phi_1,\ldots,\phi_{n-2}\in[0,2\pi)$ and one azimuthal angle $\theta \in [0,\pi/2)$. $\endgroup$ – RMWGNE96 May 13 at 14:24
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The factorisation into single integrals is fine, but for the finite result you do need to be told that each $\Re a_i\gt 0$. If even one $a_i$ has real part $\le 0$, say $a_1$, the $\int_0^\infty\exp(-a_1x_1)dx_1$ factor diverges.

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