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.
On
The inequalities given with the first double integral are $0\leq y \leq \infty$ and $-\infty \leq x \leq -y$.
If you plot the funtions and the regions you get the region:
As you can see, if you we let the $x$ be constant limits instead of $y$ the same region can be defined as $-\infty \leq x \leq 0$ and $0\leq y \leq -x$. Which gives you the bounds in your question.
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 $$\int_{0}^{\infty} \int_{-\infty}^{-y} f(x,y)dx\, dy =\iint_{\mathbb{R}^2} f(x,y)\Bigl[(x<0) \text{ and } (0 < y < -x )\Bigr]dx \,dy = \int_{-\infty}^0 \int_{0}^{-x} f(x,y) \,dy\,dx \,.$$