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By using Fubini/Tonelli Theorems, evaluate
$$\int^{0}_{1} \int^{1}_{y} x^{-\frac{3}{2}}\cos\left(\frac{\pi y}{2x}\right)dxdy$$

My attempt: by Tonelli Thm $$\int^{0}_{1} \int^{1}_{y} \left|x^{-\frac{3}{2}}\cos\left(\frac{\pi y}{2x}\right)\right|dxdy = \int^{0}_{1}\frac{dx}{x^{3/2}} \int^{x}_{0} \left|\cos\left(\frac{\pi y}{2x}\right)\right|dy $$ (at this stept I am not sure about endpoints) $$\int^{0}_{1}\frac{dx}{x^{3/2}} \int^{x}_{0} \left|\cos\left(\frac{\pi y}{2x}\right)\right|dy = \int^{0}_{1}\frac{dx}{x^{-3/2}}\times \left|\sin\left(\frac{\pi y}{2x}\right)\right|^x_0 \frac{2x}{\pi} =\frac{2}{\pi} \int^{0}_{1}\frac{dx}{x^{1/2}} = \frac{4}{\pi} $$

Is this solution true?

I think, the integral of the absolute value part is incorrect, but I couldn't do anything more than this solution. Can you help me to improve my solution?

Moreover, Tornelli Thm says us that, for nonnegative and measurable $f$, $$\int_A \int_B f=\int_B\int_Af.$$

But for this example, one end point is $y$ so I am confused a little

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Your integrand is dominated by the (positive) function $x^{-3/2}$; using Tonelli, $$ \int_0^1\int_y^1 x^{-3/2}\,dx\,dy =\int_0^1\int_0^x x^{-3/2}\,dy\,dx=\int_0^1 x^{-1/2} = 2<\infty. $$ Consequently, Fubini can be applied to your original integrand: $$ \eqalign{ \int_0^1\int_y^1 x^{-3/2}\cos(\pi y/2x)\,dx\,dy &=\int_0^1\int_0^x x^{-3/2}\cos(\pi y/2x)\,dy\,dx,\cr } $$ which you've shown how to evaluate.

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  • $\begingroup$ You mean, first part says us this function is integrable? then we can use Fubini? $\endgroup$ – corcia candy Jan 17 '16 at 17:53
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    $\begingroup$ That is correct. $\endgroup$ – John Dawkins Jan 18 '16 at 17:18

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