= Solution
{c}
The inverse <change of variables> is $x=v/(1+u)$, $y=uv/(1+u)$. Thus the triangular region becomes
$$
0\le u\le1,\qquad 0\le v\le1+u.
$$
The upper limit depends on $u$: the new region is not a <rectangle>. The absolute inverse <Jacobian determinant> is $x^2/(x+y)$, so the <change of variables formula> cancels the rational factor in the <integrand>:
$$
\begin{aligned}
I&=\int_0^1\int_0^{1+u}e^v\,dv\,du
=\int_0^1(e^{1+u}-1)\,du\\
&=\boxed{e^2-e-1.}
\end{aligned}
$$
Although the original <integrand> is undefined at the vertex $x=y=0$, its nonnegative <improper integral> exists. One can apply the <change of variables formula> on $x\ge\varepsilon$ and let $\varepsilon\downarrow0$ by the <monotone convergence theorem>; the transformed <integrand> has the finite <integral> just computed.
Back to article page
