# Breaking up a double integral product? Is this some special property?

I've been tasked to solve for $$a$$ in the following double integral equation:

$$1=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}ae^{-(x1+x2)}dx_{1}dx_{2}$$

When referencing the provided solution, my professor makes the following jump:

$$1=a\int_{-\infty }^{\infty }e^{-x1}dx_{1}\int_{-\infty }^{\infty }e^{-x2}dx_{2}$$

I'm curious as to why we the equation can be formatted into this form. I understand that the exponent addition can be transformed into a product, but why are you able to split up the double integral into a product? Is this some property of double integrals?

## 2 Answers

We have $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} ae^{-x_1+x_2}dx_1dx_2 = a\int_{-\infty}^{\infty} e^{-x_2}\underbrace{\int_{-\infty}^{\infty} e^{-x_1}dx_1}dx_2,$$ now note that the entire underbraced integral is some constant with respect to the $$x_2$$ integral, so it may be factored out:

$$a\underbrace{\int_{-\infty}^{\infty} ae^{-x_1}dx_1}\int_{-\infty}^{\infty} e^{-x_2}dx_2.$$

Yes. Double integrals can be re-written as iterated integrals, you can choose either order. In this case, the domain is the whole plane, and the function has a special structure $$f(x,y)=g(x)h(y)$$. then $$\iint_{\mathbb{R}^2}f(x,y)dxdy=\int\limits_{-\infty}^{\infty}g(x)(\int\limits_{-\infty}^{\infty}h(y)\,dy)\,dx=\int\limits_{-\infty}^{\infty}g(x)\,dx\cdot\int\limits_{-\infty}^{\infty}h(y)\,dy.$$