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


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?


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 $$ \iint_{\mathbb{R}^2}f(x,y)dxdy=\int\limits_{-\infty}^{\infty}g(x)(\int\limits_{-\infty}^{\infty}h(y)\,dy)\,dx=\int\limits_{-\infty}^{\infty}g(x)\,dx\cdot\int\limits_{-\infty}^{\infty}h(y)\,dy. $$


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