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How would I go about solving the following double integral?

$\int_{0}^{1}\int_{0}^{1-y} \sin\frac{x-y}{x+y}\mathrm dx\mathrm dy$

I am absolutely clueless on what to do with that sine.

Amir Omidi
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    Have you tried to do a $u=x-y$, $v=x+y$ substitution (haven't tried myself, just a question)? – Martigan Jun 12 '15 at 11:12
  • I tried it, I'm going to try it again to see what happens. – Amir Omidi Jun 12 '15 at 11:15
  • @Martigan it feels like I'm going in a loop, doesn't look like it's going to be helpful. – Amir Omidi Jun 12 '15 at 11:29
  • maybe related: http://math.stackexchange.com/questions/787926/calculating-trigonometric-integral https://math.stackexchange.com/questions/2313378/integral-int-01-int-01-y-cos-left-fracx-yxy-rightdxdy https://math.stackexchange.com/questions/2452462/how-to-get-the-interval-after-change-of-variables – AsukaMinato Jul 11 '20 at 03:13

1 Answers1

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Our integral is the integral of $\sin\frac{x-y}{x+y}$ over the triangle with vertices in the origin, $(0,1)$ and $(1,0)$. By rotating this region $45^\circ$ clockwise around the origin, i.e. by setting $u=\frac{x+y}{\sqrt{2}},v=\frac{x-y}{\sqrt{2}}$, we get:

$$ I = \int_{0}^{\sqrt{2}}\int_{-u}^{u}\sin\frac{v}{u}\,dv\,du=\int_{0}^{\sqrt{2}}u\int_{-1}^{1}\sin t\,dt\,du = 0.$$

Jack D'Aurizio
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