# Integrating the following function

I am asked to evaluate the integral $$\int_{R^n}e^{-\sum_{i=1}^na_i^2x_i^2}\,dx$$ given that $a_1,a_2,\cdots,a_n$ are real numbers different from $0$.

I am clueless on how to approach this, since I have never been exposed to these types of integrals. Any help is appreciated!

• Your integral is equal to $\prod_{i=1}^n \int_{\mathbb{R}} e^{-a_i^2x_i^2} dx_i$. – Hanul Jeon Mar 18 '18 at 8:06
• How do you notice that? – user372003 Mar 18 '18 at 8:09
• @user372003 It's an application of Fubini's theorem – rubik Mar 18 '18 at 9:30
• @rubik: I mean, Fubini's theorem is not how you notice it, just how you prove that what you noticed is correct. – Mehrdad Mar 19 '18 at 2:47

The function is measurable because it is continuous and, also, positive. By the Fubini-Tonelli Theorem, \begin{aligned}\int_{\mathbb{R}^n}\exp\left({-\sum_{i=1}^na_i^2x_i^2}\right)\,dx&=\prod_{i=1}^n\int_{\mathbb{R}}e^{-a_i^2x_i^2}\,dx_i=\prod_{i=1}^n\left(\int_{\mathbb{R}}e^{-a_i^2x^2}\,dx\right)\\&=\left(\frac{1}{|a_1\cdots a_n|}\int_{\mathbb{R}}e^{-x^2}\,dx\right)^n=\frac{(\sqrt{\pi})^n}{|a_1\cdots a_n|}. \end{aligned}
Applying the previous theorem and a change of variable to $(\int_{\mathbb{R}}e^{-x^2}\,dx)^2\,$ we obtain $$\left(\int_{\mathbb{R}}e^{-x^2}\,dx\right)^2=\int_{\mathbb{R}}e^{-x^2}\,dx\int_{\mathbb{R}}e^{-y^2}\,dy=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dxdy=\int_{]0,+\infty[\times]0,2\pi[}re^{-r^2}\,drd\theta=\pi.$$ Take the square root of both sides, and the result follows.