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Evaluate $$\iint_{R} \; \ln\left(\frac{x-y}{x+y}\right)\;dxdy$$ where $R$ is the triangular region with vertices $(1,0), (4,-3)$ and $(4,1)$

My work. I tried using the transformation $u=x-y$, $v=x+y$ with $|J|=1/2$ which changed the given integral as follows: $$1/2\iint_{R_{1}} \; \ln\left(\frac{u}{v}\right) \; dudv$$ where $R_{1}$ is triangle formed with vertices $(1,1)$, $(3,5)$, $(7,1)$

But this was no help at all since the integration is still very difficult... I can't think of any other transformation as well. Please help me

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  • $\begingroup$ Maybe try splitting $ln(\frac{u}{v}) = ln(u) - ln(v)$ and then use the linearity of integrals, and set $ln(.) = t$ for each integral? $\endgroup$ Commented Nov 6, 2017 at 5:48
  • $\begingroup$ You could try a transformation that maps to a right triangle. $\endgroup$ Commented Nov 6, 2017 at 6:27
  • $\begingroup$ @learning it would still be too lengthy $\endgroup$ Commented Nov 6, 2017 at 11:38
  • $\begingroup$ @PhilipHoskins I think it needs a transformation that can be integrated with the ln term by further substitution $\endgroup$ Commented Nov 6, 2017 at 11:40

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You are on the right track. The integral now is $$\frac{1}{2}\iint_{R_1} \; \ln(u/v)\;dudv=\frac{1}{2}\int_{v=1}^5 \int_{u=(1+v)/2}^{8-v} \ln(u/v)\;dudv\\= \frac{1}{2}\int_{v=1}^5 \left[u\ln(u/v)-u\right]_{u=(1+v)/2}^{8-v}dv.$$ Can you take it from here? Actually the final result is not very pretty.

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