# Show that $\int^1_0 \int^{1-x}_0e^\frac{y}{x+y}dy\,dx=\frac{e-1}{2}$

Show that $\int^1_0 \int^{1-x}_0e^\frac{y}{x+y}dy\,dx=\frac{e-1}{2}$

This is probably very simple, but I was instructed to use the substitution $x+y=u, y=uv$.

So by manipulating the integral and calculating the Jacobian (which is simply $u$, ie. $dx\,dy=u\,du\,dv$, I can express the transformation as $$\int \int ue^v dv \,du$$

The problem is that I don't know what to plug in as the limits of integration. For example, the limit $1-x$ in the expression $\int^1_0 \int^{1-x}_0e^\frac{y}{x+y}dy\,dx$ cannot be expressed as either a function of v or u purely with the substitution given. Am I missing something here? Thanks for the help.

1. the original domain is the right triangle formed by (0,0), (1,0), (0,1) on the x-y plane
2. For $u=x+y$, consider $u$ as a new axis along the line x-y=0.For the triangle domain, $u \in [0,1]$
3. From $u=x+y$ again, if you work on the contours of $x+y$ for a few values of u, they are all straight line with slope -1. If you change u in $[0,1]$, you are moving this contour line in the triangle domain. In particular, each such contour line has end points cutting the triangle edges at $v=0$ and $v =1$.
4. So, $0\leq u\leq 1$ and $0\leq v \leq 1$ is the new domain you need.
• Opssss...I don't know why I've assumed it was a square! I fix that. – gimusi Sep 4 '18 at 16:57

Since on $x-y$ plane $x+y=k$ represents a line parallel to the line $y=-x$ we have that

• $u=x+y \implies 0\le u\le 1$

then for any $u$ we have

• $0\le y\le u\implies 0\le v\le 1$
• You made a mistake, $|J|=u-uv+uv=u$. – alans Sep 4 '18 at 16:25
• opsssss....I fix. Thanks – gimusi Sep 4 '18 at 16:28

Hint: Limits of integration are 0 and 1.