Integrate $e^{\frac{x-y}{x+y}}$ over the triangle bounded by $x=0$, $y=0$, and $y=1-x$

Using change of variables, integrate $\exp\left(\frac{x-y}{x+y}\right)$ 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 $ve^{u/v} \, dv$ which is nasty.

Any ideas?

Solution 1:

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

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 $$\frac{1}{2}e^{u/v}$$ 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 $\dfrac12\left(e-\dfrac1e\right)v$, and when this is then integrated from $0$ to $1$ we get $\dfrac12\left(e-\dfrac1e\right) \cdot \dfrac12,$ or $$\frac{e-\dfrac1e}{4}.$$