Double Integral of a piecewise function If $F(x,y)$ is defined as $F(x,y) = x+y$ when $0 < x + y < 1$ and $0$ elsewhere, then find
$$\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} F(x, y) \,dx \,dy$$.
Math note: I've not yet had a professor require that I complete piecewise double integrals. However, It seems like I might simply change my limits of integration to $a=0$, $b=1$ and use those limits for the double integration of $(x+y)\,dy \,dx$. Is this correct? 
 A: I would do it in two steps, and showing your work here is essential since the answer is $+\infty$.  The first step is to evaluate the inner integral:
$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} F(x,y) \,dx \,dy = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \left\{\begin{array}{ccc}0 & : & 1 \le x+y \\ x+y & : & 0 < x+y < 1 \\ 0 & : & x+y < 0\end{array} \right\} \;\; dx \,dy.$$
By fixing $y$ to any real number you can see that the inner integral is computed over values of $y$ such that $0<x+y<1\;\;\Longleftrightarrow\;\; 0-y<x<1-y$, which gives you,
$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \left\{\begin{array}{ccc}0 & : & 1 \le x+y \\ x+y & : & 0 < x+y < 1 \\ 0 & : & x+y < 0\end{array} \right\} \;\; dx \,dy= \int_{-\infty}^{+\infty} \int_{0-y}^{1-y} x+y \;dx \,dy.$$
The inner integral is equal to $$\int_0^1 z \,dz = \frac{1}{2}$$  You may want to waive your hands a little less in this step and show your change-of-variables, this is here to give you the idea not to provide a complete copy-and-paste answer.
Hence,
$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} F(x,y) \,dx \,dy = \int_{-\infty}^{+\infty} \frac{1}{2} \,dy = +\infty.$$
