I'm trying to compute
$$\int_D \! \exp\left(\frac{x-y}{x+y}\right) \, \mathrm{d}x \, \mathrm{d}y,$$
where $D$ is the region $0 \leq x \leq 1$, $0 \leq y \leq 1-x$, a triangle in the first quadrant.
I tried to solve it using variable change, $t=x-y, s=x+y$, we can see that $0 \leq t \leq 1$ and $ 0\leq s \leq 1$
as those are the exreme values for $t,s$.
The jacobian of this transform is $\frac{1}{2}$
So overall the integral i calculated is
$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \! e^{t/s} \, \mathrm{d}t \, \mathrm{d}s=\frac{1}{2}\int_{0}^{1} \! se^{t/s} \Bigg|_{t=0}^{t=1} \, \mathrm{d}s=\frac{1}{2} \int_{0}^{1} \! s(e^{1/s}-1) \, \mathrm{d}s$$
I don't know how to find the antiderivative of this function, but even if i did, this integral does not converge.
Where is my mistake? because I know for a fact the correct answer is $\frac{e-1}{4}$.
Your error is the region for $t$ and $s$. Namely, we don't have $0 \leq t \leq 1$, since for example you could have $x = 0$, $y = 1$ giving $t = -1$.
The correct region is $0 \leq s \leq 1$, and $-s \leq t \leq s$.