<s>[INST] double integration with a domain Given that $D = \{ 1 \leq x+y \leq 2 , y+1 \leq x \leq 3y+1 \}$ and we want to evaluate $\int \int \limits_D \frac{x+y-1}{y^2 } e^{\frac{x-1}{y}}$ and used the substation $u=x+y$ and $v = \frac{x-1}{y}$ which the answer in the book use it, and get to the integral $\int \limits_{1}^{2} du \int \limits_{1}^{3} v  e^v dv  = 3e^3-e$ the problem is that the answer in the book is $e^3-e$, is my answer correct ? and if not, how to solve the question correctly ?
 [/INST] The notations:
$$\begin{cases}u=x+y\\ v = \frac{x-1}{y} \end{cases} \Rightarrow \begin{cases} x=\frac{1+uv}{1+v}\\ y=\frac{u-1}{1+v}\end{cases}$$
The Jacobian:
$$\begin{vmatrix}x_u& x_v\\ y_u&y_v\end{vmatrix}=\begin{vmatrix}\frac{v}{1+v}& \frac{u-1}{(1+v)^2}\\ \frac1{1+v}&\frac{1-u}{(1+v)^2}\end{vmatrix}=\frac{1-u}{(1+v)^2}$$
So:
$$\int \int \limits_D \frac{x+y-1}{y^2 } e^{\frac{x-1}{y}}=\int \limits_{1}^{3} \int \limits_{1}^{2} \frac{u-1}{\left(\frac{u-1}{1+v}\right)^2}e^v\cdot \frac{1-u}{(1+v)^2}dudv = \\
\int \limits_{1}^{3} \int \limits_{1}^{2} e^vdudv =\int \limits_{1}^{3} e^vdv=e^3-e.$$
Also note:
$$\int \limits_{1}^{2} du \int \limits_{1}^{3} v  e^v dv  = \int \limits_{1}^{2} \left[ve^v|_1^3-\int \limits_{1}^{3} e^vdv\right] du= \int \limits_{1}^{2} 2e^3 du=2e^3.$$
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