What should be the limits in this integration

Using the transformation $x+y = u$, $y = uv$, show that

$$\displaystyle \int_{0}^{1}\int_{0}^{1-x} e^{\frac{y}{x+y}} \, dydx = \frac{e-1}{2}.$$

In above question. I am stuck in finding limits after transformation.

My approach : $u = x + y$, $v = \frac{y} {x+y}$

I have calculated Jacobian of transformation which is $u$.

Please help me in finding limits of integration.

Solution 1:

You have two conditions

$$0 < x < 1 \quad \text{and} \quad 0 < x+y < 1. \tag{*}$$

Now the second one directly translates to $0 < u < 1$. For $v$, notice that we also have $0 < y < 1$. So we always have $0 < v < 1$. Conversely, given any $(u, v) \in (0, 1)$ we can recover $x$ and $y$ by

$$ x = u(1-v), \quad y = uv $$

In this way, the transformation $(u, v) = (u(x, y), v(x, y)) = (x+y, \frac{y}{x+y})$ is a bijection between the triangular region $\text{(*)}$ to the unit square $(0,1)^2$:

$\hspace{2em}$

(Left: region for $(x,y)$. Right: region for $(u,v)$.) Therefore

$$ \int_{0}^{1} \int_{0}^{1-x} e^{\frac{y}{x+y}} \, dydx = \int_{0}^{1}\int_{0}^{1} u e^v \, dudv = \frac{e-1}{2}. $$