for $x \in {\mathbb{R}^n}$ $\int_{\mathbb{R}^n} {e^{-|x-y|^2}dy} = \prod_{j=1}^n {\int_{-\infty}^{+\infty}{e^{-(x_j - y_j)^2}dy_j}}$. Why?

Question

According to my textbook, for $x \in {\mathbb{R}^n}$ $\int_{\mathbb{R}^n} {e^{-|x-y|^2}dy} = \prod_{j=1}^n {\int_{-\infty}^{+\infty}{e^{-(x_j - y_j)^2}dy_j}}$. Why is it so? I understand that $e^{-|x-y|^2} = \prod_{j=1}^n e^{-(x_j - y_j)^2}$. But I can't see how the integral itself is represented as a product of integrals.

Answer 1

This is due to the fact that $e^{(x_i - y_i)^2}$ only depends on $y_i$ (and $x_i$ which is a constant in you integral). Therefore, you can simply use the linearity of the integral. With $n = 2$, it gives, \begin{align*} \int_{\mathbb{R}^2} e^{-|x - y|^2} & = \int_{\mathbb{R}}\int_{\mathbb{R}} e^{-(x_1 - y_1)^2}e^{-(x_2 - y_2)^2} \, dy_1dy_2 \textrm{ by Fubini-Tonelli},\\ & = \int_{\mathbb{R}} e^{-(x_2 - y_2)^2}\int_{\mathbb{R}} e^{-(x_1 - y_1)^2} \, dy_1dy_2\\ & \textrm{ because each } e^{-(x_2 - y_2)^2} \textrm{ is constant with respect to } dy_1,\\ & = \int_{\mathbb{R}} e^{-(x_1 - y_1)^2} \, dy_1\int_{\mathbb{R}} e^{-(x_2 - y_2)^2}dy_2\\ & \textrm{because } \int_{\mathbb{R}} e^{-(x_1 - y_1)^2} \, dy_1 \textrm{ is constant with respect to } dy_2.\\ & = \prod_{k = 1}^2 \int_{\mathbb{R}} e^{-(x_k - y_k)^2} \, dy_k \end{align*} It is the same thing for any $n \geqslant 1$.