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If we have

P($x_1,x_2,x_3$) $\propto$ $\delta_D$($x_1+x_2+x_3$-1) $\Theta(x_1)$$\Theta(x_2)$$\Theta(x_3)$,

then how to find the form of P($x_1,x_2,x_3$)?

i.e., how to calculate the integral $\int\int\int$$\delta_D$($x_1+x_2+x_3$-1) $\Theta(x_1)$$\Theta(x_2)$$\Theta(x_3)$ d$x_1$d$x_2$d$x_3$?

I have been confused about this question for a couple of days and really had no idea about it. Anyone can help me?

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Assuming $\delta_D$ is the dirac delta, you can express one variable in the triple integral in terms of the other two, and reduce the integral to a double integral. For example:

$$\int\int\int \delta_D(x_1+x_2+x_3-1)\Theta(x_1)\Theta(x_2)\Theta(x_3)\text{d}x_1\text{d}x_2\text{d}x_3$$

$$=\int\int\Theta(1-x_2-x_3)\Theta(x_2)\Theta(x_3)\text{d}x_2\text{d}x_3$$

This comes from considering the single integral

$$\int \delta_D(x_1+x_2+x_3-1)\Theta(x_1)\text{d}x_1$$

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  • $\begingroup$ Thank you very much. I have another question about integral of dirac delta function: I know that \int \delta_D(x) f(x) dx =f(0) but is \int \delta_D(x) f(x) g(x) dx =f(0)*g(0)? $\endgroup$
    – yichel
    Apr 15, 2016 at 9:52
  • $\begingroup$ Yes, that is correct, you can simply consider $h(x)=f(x)g(x)$. Then $\int \delta_D{x}h(x)dx=h(0)=f(0)g(0)$ $\endgroup$
    – Comp_Warrior
    Apr 15, 2016 at 21:23

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