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Using change of variables, integrate $e^{\frac{x-y}{x+y}}$ over the triangle bounded by $x=0$, $y=0$, and $y=1-x$.

I'm using $u=x-y$ and $v=x+y$ but running into the issue where I have to integrate something like $ve^{\frac{u}{v}}dv$ which is nasty.

Any ideas?

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Please use LaTeX. –  ᴊ ᴀ s ᴏ ɴ Nov 21 '12 at 8:50

1 Answer 1

You actually don't end up integrating $v e^\frac{u}{v}$.

The variable change $u=x-y,v=x+y$ is a linear map, and so it multiplies areas everywhere by the constant amount given by the determinant of the matrix of partials, which for this linear function are all +1 or -1. I got the multiplying factor is 2, so that you end up in fact integrating $$\frac{1}{2}e^\frac{u}{v}$$ over the triangle whose vertices are $(0,0),(1,1),(-1,1)$.

Setting this up as an iterated integral, so that we integrate over $u$ and then $v$, gives $$\int_0^1 \int_{-v}^{v} e^{u/v} du dv.$$ The value of the inside integral here is $(1/2)(e-1/e)v$, and when this is then integrated from $0$ to $1$ we get $(1/2)(e-1/e) \cdot (1/2),$ or $$\frac{e-1/e}{4}.$$

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