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

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?

• The assumption that $a_i>0$ for all $i$ cannot be missed, I think. – RMWGNE96 May 13 at 14:21
• 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 $\theta \in [0,\pi/2)$. – RMWGNE96 May 13 at 14:24

## 1 Answer

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.