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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?

StubbornAtom
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1 Answers1

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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 $$\frac{e-\dfrac1e}{4}.$$

TheSimpliFire
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coffeemath
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  • I'm sorry to ask so much later, but how did you find the bounds for $u$? I can't get them from the bounds of the initial region: ${x > 0, y > 0, x + y < 1}$. From the last one I get $0 < v < 1$, but what about $u$? – rubik Feb 14 '16 at 10:03
  • $u=x-y$ can be any difference between $x$ and $y,$ so the smallest is when $x=0,y=1$ where it is $-1,$ and the largest when $x=1,y=0$ where it is $+1.$ By the way you should check whether the multiplying factor should be $1/2$ as I put it above, or $2$ instead--- I sometimes get that mixed up. [Do we multiply by Jacobin of forward or by Jacobian of inverse function, that's what I don't always get straight. Check a calc 2 book or google change of variables in double integral.] – coffeemath Feb 15 '16 at 14:29
  • Ahh, that's the reasoning, thanks! The result is correct, I get the same. – rubik Feb 15 '16 at 21:44