Determine whether the following Lebesgue integral exists.

Question

I am asked to determine whether $$\int\limits_{[0,1]\times[0,1]} f(x,y)d\lambda^2$$ where $$f = \begin{cases} y^{\frac{-3}{2}} & 0 < x^2 <y<1 \\ -2x^{-3} & 0 < y<x^2<1 \\ 0 & \text{else.}\end{cases}$$ exists or not and if it does to calculate its value.

I think it might help to calculate the iterated integrals but I don't really know how to do this.

Answer 1

The integral exists if: $$I=\int_{[0,1] \times [0,1]} |f(x,y)| d \lambda^2 < + \infty$$ By Tonelli's theorem: $$I=\int_{[0,1]} \int_{[0,1]} |f(x,y)| dy dx$$ But with $x$ fixed you have: $$\int_0^1 |f(x,y)| dy= \int_0^{x^2} 2 x^{-3} dy+\int_{x^2}^1 y^{-\frac{3}{2}}dy=x^2 \frac{2}{x^3}+\left[-\frac{2}{y^\frac{1}{2}} \right]_{x^2}^1=\frac{2}{x}+\frac{2}{x}-2$$ but: $$\int_0^1 \left(\frac{4}{x}-2 \right) dx=+ \infty$$ so the integral does not exists.