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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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