Explain why $\big(\int_{-\infty}^{\infty}e^{-z^2/2}dz \big)^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(z^2 + u^2)/2}dzdu$

Question

I came across the following when studying a proof related to the normal distribution:

$$\left(\int_{-\infty}^{\infty}e^{-z^2/2}\ dz \right)^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(z^2 + u^2)/2}\ dz\ du$$

Is this some kind of identity? It was used as a step in the proof but I would like to know how it was arrived at?

Answer 1

\begin{align} \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-\frac{1}{2}(x^2+y^2)}dxdy &=\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}y^2}dxdy\\ &=\int^{\infty}_{-\infty}e^{-\frac{1}{2}y^2}\underbrace{\left[\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}dx\right]}_{constant \ w.r.t.y}dy\\ &=\left[\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}dx\right]\left[\int^{\infty}_{-\infty}e^{-\frac{1}{2}y^2}dy\right]\\ &=\left[\int^{\infty}_{-\infty}e^{-\frac{1}{2}x^2}dx\right]^2 \end{align}