I am trying to evaluate this integral:

$$\int_{0}^{1}\int_{x}^{1} y^2 \sin(2\pi \frac{x}{y})dydx$$ $$=\int_{0}^{1}\int_{0}^{1} \chi_{[x,1]}(y) y^2 \sin(2\pi \frac{x}{y})dydx$$ $$=\int_{0}^{1}\int_{0}^{1} \chi_{[x,1]}(y) y^2 \sin(2\pi \frac{x}{y})dxdy$$ $$=\int_{0}^{1}\int_{0}^{1} \chi_{[0,y]}(x) y^2 \sin(2\pi \frac{x}{y})dxdy$$

so I am stuck here and I don't know what to do?


A first integration with respect to $x$ is clearly easier. I suggest you to apply Fubini theorem and reverse the order of integrals.

$$\int_{0}^{1}\int_{x}^{1} y^2 \sin(2\pi \frac{x}{y})dydx = \int_{0}^{1} y^2\left( \int_{0}^{y} \sin(2\pi \frac{x}{y})dx \right)dy.$$ And $$\int_{0}^{y} \sin(2\pi \frac{x}{y}) dx = \frac{y}{2\pi} [-\cos(2\pi \frac{x}{y})]_0^y = 0$$

  • $\begingroup$ why did you take $y^2$ common out in first integration? have you changed the order? $\endgroup$ – Bhaskara-III Feb 4 '16 at 22:17

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