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,

$$\tag{1} I^2 = \left( \int\limits_{-\infty}^{\infty} \normalsize e^{\small -\frac{y^2}{2}}\ dy\ \right) \left( \int\limits_{-\infty}^{\infty} \normalsize e^{\small -\frac{x^2}{2}}\ dx\ \right) = \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} e^{-(y^2 +x^2)/2}\ dy\ dx$$

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, $\forall x, y \in \mathbb{R}$ ? $$\tag{2} I^2 = \left( \int\limits_{-\infty}^{\infty} \normalsize f_y(y)\ dy\ \right) \left( \int\limits_{-\infty}^{\infty} \normalsize f_x(x)\ dx\ \right) = \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} f_y(y)f_x(x) dy\ dx$$

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.

• If $\displaystyle \left( \int_{-\infty}^{\infty} \normalsize f_y(y)\ dy\ \right)$ converges, it is a constant, let's call it $a$. Then \begin{align} \left( \int_{-\infty}^{\infty} \normalsize f_y(y)\ dy\ \right) \left( \int_{-\infty}^{\infty} \normalsize f_x(x)\ dx\ \right)&=a\left( \int_{-\infty}^{\infty} \normalsize f_x(x)\ dx\ \right)\\ &=\left( \int_{-\infty}^{\infty} a\normalsize f_x(x)\ dx\ \right)\\&=\int_{-\infty}^{\infty} \left(\int_{-\infty}^{\infty} f_y(y) dy\right)f_x(x) dx, \end{align}for similar reasons you can now get the $f_x(x)$ inside the integral on $y$. Commented Nov 3, 2013 at 9:06
• Integrals are essentially sums (in view of Riemann sum). Thus it is not so bad to say that the questioned identity holds because $$(a+b)(c+d) = a(c+d)+b(c+d) = (ac+ad)+(bc+bd).$$ Commented Nov 3, 2013 at 9:10
• We must be looking at the exact same textbook. I saw this in the proof that the normal distribution is a pdf, searched for "multiplying two integrals together," and landed right here.
– Max
Commented Dec 4, 2019 at 23:43
• Read Fubini's Theorem. Try either Wikipedia or Wolfram MathWorld for quick easy read. Commented Mar 31, 2020 at 13:44
• @GitGud what are those "similar reasons"? I don't get it Commented Jul 15, 2022 at 12:14

Note first, that for any integrable $f \colon \def\R{\mathbb R}\R \to \R$ and any $\alpha \in \R$ we have $$\int_\R \alpha f(x)\, dx = \alpha \int_\R f(x) \, dx$$ Now note that with $\alpha := \int_\R f(y)\, dy$ (this depends on $f$, but given $f$ it is a constant) this gives $$\int_\R \left(\int_\R f(y)\, dy\right)\, f(x)\, dx = \int_\R f(y)\, dy \cdot \int_\R f(x)\, dx$$ Now, for every fixed $x \in \R$, we apply the above again, now for $\alpha = f(x)$ (which is not depending on $y$), giving $$\left(\int_\R f(y)\, dy\right)\, f(x) = \int_\R f(x)f(y)\, dy$$ altogether $$\int_\R \int_\R f(y)f(x)\, dy\, dx = \int_\R f(y)\, dy \cdot \int_\R f(x)\, dx$$
• Because $\int f(y) dy$ is a constant with respect to x, and you can use the first line to move it outside the integral with respect to x. Commented Oct 24, 2020 at 5:25
• @Dargscisyhp To be pedantic, the integral IS a constant (not just a constant with respect to $x$) Commented Dec 28, 2020 at 0:54
• @masiewpao Sure. But to do what we've done, it's sufficient to be independent of $x$. And any constant is especially constant with respect to $x$. Commented Dec 28, 2020 at 12:20