# How the product of two integrals is iterated integral? ∫⋅∫=∬\int\cdot \int = \iint

Let $\large{I} = \Large \int\limits_{-\infty}^{\infty} \normalsize e^{\small -\frac{y^2}{2}}\ dy$.

Then, my textbook says,

I am not seeing how we reach the rightmost, iterated integral, and I do not remember the calculus how or why this works. Why can we restate the product of two integrals, each having funcions of different variable, as an iterated integral including both variables?

Let $\large{I} = \Large \int\limits_{-\infty}^{\infty} \normalsize f_y(y)\ dy$ for any continuous, differentiable $f_y: \mathbb{R} \to \mathbb{R}$

Now, does (2) generally hold, $\forall x, y \in \mathbb{R}$ ?

or is result(1) specific to the particular function $e^{-(y^2)/2}$?

Please show the work or reasoning behind how and why this transformation is valid.

Note first, that for any integrable $f \colon \def\R{\mathbb R}\R \to \R$ and any $\alpha \in \R$ we have
Now note that with $\alpha := \int_\R f(y)\, dy$ (this depends on $f$, but given $f$ it is a constant) this gives
Now, for every fixed $x \in \R$, we apply the above again, now for $\alpha = f(x)$ (which is not depending on $y$), giving