Compute an improper integral.

Question

Suppose $A = [0,\infty) \times [0, \infty) $. Let $f(x,y) = (x+y)e^{-x-y} $. How can I find $ \int_A f $? I know since $f$ is continuous on $A$, then $\int _A f $ exists, Do I need to evaluate

$$ \int_{0}^{\infty} \int_{0}^{\infty} (x+y)e^{-x-y}\ dx\ dy\ ??$$

Answer 1

Let $A_M$ be the set $[0,M]\times [0, M]$ and $I_M=\int_{A_M}{f}$. From Fubini's Theorem \begin{align} I_M&=\int_{y=0}^{y=M}\left[\int_{x=0}^{x=M}(x+y)e^{-(x+y)}dx\right]dy \\ &=\int_{y=0}^{y=M}\left[-(x+y+1)e^{-(x+y)}\right]_{x=0}^{x=M}dy \\ &=\int_{y=0}^{y=M}\left[(y+1)e^{-y}-(M+y+1)e^{-(M+y)}\right]dy \\ &=\left[-(y+2)e^{-y}+(M+y+2)e^{-(M+y)}\right]_{y=0}^{y=M} \\ &=-(M+2)e^{-M}+2+2(M+1)e^{-2M}-(M+2)e^{-M} \\ &=2-2(M+2)e^{-M}+2(M+1)e^{-2M} \end{align} Now, \begin{align} \int_A{f}&=\lim_{M\rightarrow \infty}{I_M} \\ &=\lim_{M\rightarrow \infty}{\left[2-2(M+2)e^{-M}+2(M+1)e^{-2M}\right]} \\ &=2. \end{align}

Answer 2

HINT :

Rewrite: $$ \int_{0}^{\infty} \int_{0}^{\infty} (x+y)e^{-x-y} dx dy =\int_{0}^{\infty} e^{-y} \int_{0}^{\infty} (x+y)e^{-x} dx dy\tag1 $$ Note that $$ \int_0^\infty z^n e^{-z}\ dz=(n+1)!\tag2 $$ Use $(2)$ to evaluate $(1)$.