0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $$\int_{0}^{y} \sin(2\pi \frac{x}{y}) dx = \frac{y}{2\pi} [-\cos(2\pi \frac{x}{y})]_0^y = 0$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.