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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.

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    Ah I know that so it works very well thanks.2012-06-12

1 Answers 1

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\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}

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    Thanks for your kind answer. I got it.2012-06-12