multivariable calculus - Changing order of integration:$int_0^inftyint_{-infty}^{-y}f(x)mathrm dxmathrm dyRightarrowint_{-infty}^0int_0^{-x}f(x)mathrm dymathrm dx$

Why does

$$\int_{0}^{\infty} \int_{-\infty}^{-y} f(x)\mathrm dx \mathrm dy \Rightarrow \int_{-\infty}^{0} \int_{0}^{-x} f(x) \mathrm dy \mathrm dx$$

The title is pretty self explanatory. I couldn't see how to properly change the order of the left integeral to the right one.

I'd love to hear your thoughts, thanks.

Answer

The easiest way to perform a change of the order of integration in the multivariable setting is via Iverson's bracket. This is the indicator function such that

$$[P] =\begin{cases} 1 & P \text{ is true}\\ 0 & \text{else}.\end{cases}$$ The change of variable arises from reinterpreting the system of inequalities (see below).

With the Iverson notation, one can remove the boundaries from the integral and implement it in the integrand, i.e.,

$$\int_{0}^{\infty} \int_{-\infty}^{-y} f(x,y)dx \,dy =\iint_{\mathbb{R}^2} f(x,y)\Bigl[(y\geq 0) \text{ and } (x\leq-y) \Bigr]dx\, dy \,.$$

Now in order to perform the change of the order of integration, you have to reinterpret Iverson's bracket. You have to figure out what condition

$$P=(y\geq0) \text{ and } (x\leq-y)$$ poses on $x$ first.

The maximal value that $x$ can achieve is $0$ (when $y=0$). The second condition in $P$ is equivalent to

$$x \leq -y \Leftrightarrow y \leq -x \,.$$

The first condition demands that $y>0$. Together, we have that
lf that $P$ is equivalent to
$$P \Leftrightarrow (x<0) \text{ and } (0 < y < -x )\,.$$

So we find

$$\int_{0}^{\infty} \int_{-\infty}^{-y} f(x,y)dx\, dy =\iint_{\mathbb{R}^2} f(x,y)\Bigl[(x<0) \text{ and } (0 < y < -x )\Bigr]dx \,dy = \int_{-\infty}^0 \int_{0}^{-x} f(x,y) \,dy\,dx \,.$$