# $\int_E \frac{y^2}{x^2}d\lambda^2(x,y)$ for $E=\{ (x,y)\in\mathbb R^2 : x\le y, xy\ge 1, y\le2 \}$

$\int_E \frac{y^2}{x^2}d\lambda^2(x,y)$ for $E=\{ (x,y)\in\mathbb R^2 : x\le y, xy\ge 1, y\le2 \}$

I often struggle to find the boundaries of integration from a given area.

My approach was to use Fubini as the function is positive and compute the iterated integral

$\int_{-\infty}^{2}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy=...=\infty$

My questions is whether the boundaries $\frac{1}{y} \le x\le y$ and $-\infty \le y\le 2$ are correct and if there is a general/elegant approach to find them from a given arbitrary area.

• See that while $y \in (-\infty , -1) \cup (0,1)$, we have that $\frac{1}{y} > y$, so this parts should be discarded. Also note that $y = 0$ also can't be the situation. So in general $y \in [1, 2] \cup (-1,0)$. – kolobokish Feb 18 '17 at 22:14
• I meant in previous comment the sitation where you write $\int_{\frac{1}{y}}^{y}$. And for the case where $y\in (\infty, -1) \cup (0,1)$, it should be written $\int_{y}^{\frac{1}{y}}$. – kolobokish Feb 18 '17 at 22:24
• @kolobokish So $\int_E \frac{y^2}{x^2}d\lambda^2(x,y)=\int_{1}^{2}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy+\int_{-1}^{0}\int_{1/y}^{y} \frac{y^2}{x^2} dx dy+\int_{-\infty}^{-1}\int_{y}^{1/y} \frac{y^2}{x^2} dx dy+\int_{0}^{1}\int_{y}^{1/y} \frac{y^2}{x^2} dx dy$ ?? – Blablablu Feb 18 '17 at 22:35
• Yes. It seem to me, it must be so. – kolobokish Feb 18 '17 at 22:40

Wolfram Alpha produces this plot for x <= y, xy >= 1, y <= 2:
So it makes sense that you would need to break the integration up into at least two pieces. Since $xy > 0$, $x$ and $y$ are always of the same sign.
When $x > 0, y> 0$, then $y = 2$ is the top, $y = x$ is bottom and left, and $y = 1/x$ is bottom and right. $y = 1/x$ and $y = x$ cross at $(1,1)$, so if you integrate by $y$ first, you need $y \in [1/x, 2]$ for $x \le 1$ and $y \in [x, 2]$ for $x \ge 1$. $y = 1/x$ crosses $y = 2$ at $x = 1/2$, so you get $\int_{1/2}^1\int_{1/x}^2 \, dydx$ for the first and $\int_1^2\int_{x}^2 \, dydx$ for the second.
But if you integrate $x$ first, then this piece can be done in a single integration: $\int_1^2\int_{1/y}^y \, dxdy$
The lower left piece where $x < 0, y < 0$ is simpler, as $y \le 1/x$ implies $y \le 2$ here. The integral is $\int_{-\infty}^{-1}\int_{x}^{1/x}\, dydx$
So the simplest form is $$\int_E \frac{y^2}{x^2}d\lambda^2(x,y) = \int_{-\infty}^{-1}\int_{x}^{1/x}\frac{y^2}{x^2}\,dydx + \int_1^2\int_{1/y}^y \frac{y^2}{x^2}\, dxdy$$