# 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

By Fubini, \begin{align*} I&=\int_{{\bf{R}}^{n}}e^{-a_{1}^{2}x_{1}^{2}}\cdots e^{-a_{n}^{2}x_{n}^{2}}dx_{1}\cdots dx_{n}\\ &=\int_{{\bf{R}}}\cdots\int_{\bf{R}}e^{-a_{1}^{2}x_{1}^{2}}dx_{1}\cdots e^{-a_{n}^{2}x_{n}^{2}}dx_{n}\\ &=\left(\int_{\bf{R}}e^{-a_{1}^{2}x_{1}^{2}}dx_{1}\right)\cdots\left(\int_{{\bf{R}}}e^{-a_{n}^{2}x_{n}^{2}}dx_{n}\right)\\ &=\dfrac{\sqrt{\pi}}{|a_{1}|}\cdots\dfrac{\sqrt{\pi}}{|a_{n}|}. \end{align*}

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}

In case you are not familiar with the Gaussian Integral, here is a quick result.

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.