Please help understand why the square $I^2$ of the integral of exponential$$
I = \int_{-\infty}^{\infty} exp(\frac {-1}{2\sigma^2} \, x^2) \, dx \tag{1}
$$
can be expressed as double integral as in (2).
$$
\begin{align}
I^2 &=
\int_{-\infty}^{\infty} exp(\frac {-1}{2\sigma^2} \, x^2) \, dx
\int_{-\infty}^{\infty} exp(\frac {-1}{2\sigma^2} \, y^2) \, dy
\\
&=
\iint_{-\infty}^{\infty}
exp \Bigl( \frac {-1}{2\sigma^2} \, (x^2 + y^2) \Bigl) \, \, dx \, dy
\tag{2}
\end{align}
$$
I believe the product of integrals can be double integral because of exponential but could not understand how it can be transformed as such.
Pattern Recognition and Machine Learning
Best Answer
If the function $f$ can be separated into the product of two functions $X$ and $Y$ - where each of them only rely on $x$ and $y$ respectively - you can split the double integral like that.
$$\iint f(x, y) dy dx = \iint X(x) Y(y) dydx = \int X(x) \int Y(y) dy dx = \int X(x) dx \int Y(y) dy$$
since $X(x)$ and $dx$ are all constants when we integrate $Y(y) dy$. $X(x) = ce^{-x^{2}}$ and $Y(y) = ce^{-y^{2}}$ is your case.
When the integral is squared, there should be two $x$'s then change one into $y$ (Because the name of the parameters does not matter).