evaluating $\int_0^\infty \int_y^\infty y^2e^{-x^4} \ dx \ dy$ evaluating $\int_0^\infty \int_y^\infty y^2e^{-x^4} \ dx \ dy$
my book states  $$\int_0^\infty \int_y^\infty y^2e^{-x^4} \ dx \ dy = \int_0^\infty \int_0^x y^2e^{-x^4} \ dy \ dx$$
Could someone explain how the domains have changed?
 A: $\displaystyle\int_0^\infty \left( \int_y^\infty \cdots \,dx\right) \,dy$ means $y$ is a positive number, and for any particular value of $y$, $x$ must be bigger than $y$.
$\displaystyle\int_0^\infty \left( \int_0^x \cdots\,dy\right) \,dx$ means $x$ is a positive number, and for any particular value of $x$, $y$ must be smaller than $x$ but still positive.
They both say $0 < y < x < \infty$.
The second form, $\displaystyle\int_0^x y^2e^{-x^4} \ dy$ is tractable because:


*

*first you observe that as $y$ goes from $0$ to $x$, the factor $e^{-x^4}$ does not change, so it can be written as $$ e^{-x^4} \int_0^x y^2\,dy,$$ and

*next you observe that evaluation of that remaining integral is trivial, and

*then you're left with a function of $x$ that yields immediately to the subsitution $u = x^4$.

A: The interchanging of two $\int$ signs is justified using the Fubini theorem. Moreover, notice that
$$(x,y)\in [y,\infty)\times[0,\infty)\iff 0\le y\le x\iff (x,y)\in[0,\infty)\times[0,x]$$
and this explains how the domain has changed.
