Is $f(x,y)=\operatorname{sgn}(x-y)e^{-|x-y|}$ Lebesgue integrable? Let $f(x,y)=\operatorname{sgn}(x-y)e^{-|x-y|}$
a) Compute $\int_{0}^{\infty}\int_{0}^{\infty} f(x,y)\,dy\,dx$ and $\int_{0}^{\infty}\int_{0}^{\infty}f(x,y)\,dy\,dx$.
b) Is $f$ Lebesgue integrable on $(0,\infty)\times (0,\infty)$?
I have this: 
For a) If $x\geq y,\ \int_{0}^{\infty}\int_{0}^{\infty} f(x,y)\,dx\,dy=\int_{0}^{\infty}\int_{y}^{\infty} e^{-(x-y)}\,dx\,dy=\int_{0}^{\infty}e^{y} \int_{y}^{\infty} e^{-x}\,dx\,dy=\int_{0}^{\infty}e^{y}e^{-y}\,dx=\infty$.
Similarly, for $x<y$, $\int_{0}^{\infty}\int_{0}^{y} e^{x-y}\,dy\,dx=\infty$.
In a similar way, the other integral is calculated.
For b)  As an integral gave us infinite then the function would not be integrable?
 A: Recall that
$$ (|u|)^\prime=\frac{|u|}{u}\cdot u^\prime=\operatorname{sgn}(u)\cdot u^\prime\text{ for }u\ne0 $$
So
\begin{eqnarray}
\frac{\partial}{\partial y}e^{-|x-y|}&=& e^{-|x-y|}\frac{\partial}{\partial y}(-|x-y|)\\
&=&-\operatorname{sgn}(x-y)\frac{\partial}{\partial y}(x-y)e^{-|x-y|}\\
&=&-\operatorname{sgn}(x-y)\cdot(-1)\\
&=&\operatorname{sgn}(x-y)e^{-|x-y|}\end{eqnarray}
Therefore
$$ \int\operatorname{sgn}(x-y)e^{-|x-y|}\,dy=e^{-|x-y|}+c $$
If we define
$$ f(x,y)=\begin{cases}\operatorname{sgn}(x-y)e^{-|x-y|}&\text{ for }y\ne x\\
1&\text{ for }x=y\end{cases} $$
then
\begin{eqnarray}
\int_{0}^{\infty}\int_{0}^{\infty} f(x,y)\,dy\,dx&=&\int_0^\infty e^{-|x-y|}\Big|_0^\infty\,dx\\
&=&\int_0^\infty-e^{-x}\,dx\\
&=&1
\end{eqnarray}
Note that $\left\{(x,y)\,|\,x=y\right\}$ is a set of measure $0$.
**ADDENDUM **
To evaluate the integral by breaking it into two parts we can proceed as follows:
\begin{eqnarray}
\int_0^\infty\int_0^\infty f(x,y)\,dy\,dx&=&\int_0^\infty\left(\int_0^x f(x,y)\,dy+\int_x^\infty f(x,y)\,dy\right)\,dx\\
&=&\int_0^\infty\left(\int_0^x e^{y-x}\,dy-\int_x^\infty e^{x-y}\,dy\right)\,dx\\
&=&\int_0^\infty\left(e^{y-x}\Big|_0^x+e^{x-y}\Big|_x^\infty\right)\,dx\\
&=&\int_0^\infty(-e^{-x})\,dx\\
&=&1
\end{eqnarray}
