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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.

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mon
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    A definite integral returns a number, no matter what variable is used. So writing, say $e^{-x^2}$ or $e^{-y^2}$ in the integrand doesn't matter. – Sean Roberson Jan 02 '23 at 04:32

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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).

Vue
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