By using the transformation x+y=u. y=uv show that the integral e^y/(x+y) dydx = (e-1)/2 with x goes from 0 to 1, and y goes from 0 to 1-x.

p/s: I am sorry. I am so bad in computer. I can not post the image of integral in this post.

Is your integral \displaystyle \begin{align*} \int_0^1{\int_0^{1-x}{\frac{e^y}{x + y}\,dy}\,dx} \end{align*}? Have you started by evaluating the Jacobian? \displaystyle \begin{align*} J = \left[ \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right] \end{align*}

Originally Posted by Prove It
Is your integral \displaystyle \begin{align*} \int_0^1{\int_0^{1-x}{\frac{e^y}{x + y}\,dy}\,dx} \end{align*}? Have you started by evaluating the Jacobian? \displaystyle \begin{align*} J = \left[ \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right] \end{align*}
yes, this is my integral. I have already calculated the determinant. However, I had to stop my work because I have a problem when I changed (x,y) to (u,v). Please help me.