Prove equality of integrals Let $f(x,y) = \text{sgn}(x-y)e^{-|x-y|}$ (Where $\text{sgn}(t)$ is the sign of $t$)
I want to prove the equation below.
$$\int^\infty_0dx \int^\infty_0 f(x,y)dy = -\int^\infty_0 dy \int^\infty_0 f(x,y)dx =-1$$
I don't know how can I start to prove this. Please give some outline for that.
 A: \begin{align}
\int_0^{\infty} f(x,y) dy & = \int_0^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dy\\
& = \int_0^{x} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dy + \int_x^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dy\\
& = \int_0^{x} \exp(-\lvert x - y\rvert) dy - \int_x^{\infty} \exp(-\lvert x - y\rvert) dy\\
& = \int_0^{x} \exp(-(x - y)) dy - \int_x^{\infty} \exp(x-y) dy\\
& = \int_0^{x} \exp(y-x) dy - \int_x^{\infty} \exp(x-y) dy\\
& = \left. \exp(y-x) \right \rvert_{0}^{x} + \left. \exp(x-y) \right \rvert_{x}^{\infty}\\
& = \left(1 - \exp(-x) \right) + \left( 0 - 1\right)\\
& = - \exp(-x)
\end{align}
Hence,
\begin{align}
\int_0^{\infty} \left(\int_0^{\infty} f(x,y) dy \right) dx & = \int_0^{\infty} - \exp(-x) dx = \left. \exp(-x) \right \rvert_{0}^{\infty} = 0 - 1 = -1
\end{align}
You can do a similar thing for $$\int_0^{\infty} \left(\int_0^{\infty} f(x,y) dx \right) dy$$
\begin{align}
\int_0^{\infty} f(x,y) dx & = \int_0^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dx\\
& = \int_0^{y} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dx + \int_y^{\infty} \text{sgn}(x-y) \exp(-\lvert x - y\rvert) dx\\
& = \int_0^{y} -\exp(-\lvert x - y\rvert) dx + \int_y^{\infty} \exp(-\lvert x - y\rvert) dx\\
& = \int_0^{y} -\exp(x - y) dx + \int_y^{\infty} \exp(y-x) dx\\
& = -\left. \exp(x-y) \right \rvert_{0}^{y} - \left. \exp(y-x) \right \rvert_{y}^{\infty}\\
& = - \left(1 - \exp(-y)\right) - \left( 0 - 1\right)\\
& = \exp(-y)
\end{align}
Hence,
\begin{align}
\int_0^{\infty} \left(\int_0^{\infty} f(x,y) dx \right) dy & = \int_0^{\infty} \exp(-y) dy = - \left. \exp(-y) \right \rvert_{0}^{\infty} = - \left(0 - 1 \right) = 1
\end{align}
