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I am trying to understand the justification of step (*) below, which I have seen used to find the integral of the Error Function Er(f); the cdf of the standard-normal normal .

Here is the derivation I know:

We set $I:= \int_{-\infty}^\infty e^{-x^2}dx$

Then we use: $$I^2= \left( \int_{-\infty}^\infty e^{-x^2} dx\right) \left(\int_{-\infty}^\infty e^{-y^2} dy\right)$$

Then we end up with: (*) $$I^2=\int_{-\infty}^\infty e^{-(x^2+y^2)} dxdy$$ --and we have a nice polar integral. question:

How do we conclude that $\big(\int f(x)dx\big) \big(\int f(y)dy\big) = \int f(x)f(y) dxdy$?

AFAIK , integration is not multiplicative. What gives?

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What you really have in the last step is a double integral, call it $J$: $$J = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x) f(y)\,dx\,dy,$$ which really means $$J = \int_{-\infty}^\infty \left( \int_{-\infty}^\infty f(x) f(y)\,dx \right) \,dy.$$ Because the inner integral is over $x$, the factor of $f(y)$ is a constant and can be pulled out, and you get $$\begin{align}J &= \int_{-\infty}^\infty f(y) \left( \int_{-\infty}^\infty f(x)\,dx \right) \,dy \\ &= \int_{-\infty}^\infty f(y) I \,dy. \\ &= I \int_{-\infty}^\infty f(y) \,dy \\ &= I \cdot I \\ &= I^{2}.\end{align}$$

Note that integration is not multiplicative in the sense that $\int f(x) g(x) dx \neq \left(\int f(x) dx\right) \left(\int g(x) dx\right)$. However, it is true that $\iint f(x) g(y) dx dy = \left(\int f(x) dx\right) \left(\int g(y) dy\right)$.

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This can be justified by Fubini's Theorem.

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