# Changing the Order of Integration, Integrability

Suppose I have the integral $$\int_{0}^{1} \int_{y}^{1} x^{-3/2} \cos(\frac{\pi y}{2x}) dx dy$$. I know that if I change the order of integration I get $$\frac{4}{\pi}$$. But how do I apply Fubini's theorem to do so (or can I?). I know if $$\left| \int_{y}^{1} f(x,y) dy \right| < \infty$$, then I can change the order, but I am not sure how to proceed.

Put the absolute value to the integrand, and you just need to check that \begin{align*} \int_{0}^{1}\int_{y}^{1}x^{-3/2}dxdy<\infty. \end{align*}
Well we know that for your $$f(x,y)$$, $$|\int_{0}^{1} \int_{y}^{1} f(x,y) dx dy| \leq \int_{0}^{1} \int_{y}^{1} |f(x,y)| dx dy \leq \int_{0}^{1} \int_{0}^{1} |f(x,y)| dx dy$$. If this is integrable, then you are golden.
Note that $$|\cos{(...)}| \leq 1$$
Also so you have $$|f(x,y)|=|\cos{(...)}\frac{1}{x^{\frac{3}{2}}}1_{(y,1]}(x)| \leq \frac{1}{x^{\frac{3}{2}}}1_{(y,1]}(x)$$
Now $$\int_0^1 \int_0^1|f(x,y)|dxdy=C\int_0^1 1-\frac{1}{\sqrt{y}}dy<+\infty$$