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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.

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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*}

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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.

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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 $\int_0^1 \int_0^1|f(x,y)|dxdy=C\int_0^1 1-\frac{1}{\sqrt{y}}dy<+\infty$

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