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Anybody could help with some demonstration?

$ \int_0^a \int_0^a e^{ik_1x_1} e^{ik_2x_2} e^{-q|x_1-x_2|} dx_1dx_2 $

lorniper
  • 225

2 Answers2

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$$\begin{gather}\int_0^a \int_0^a e^{ik_1x_1} e^{ik_2x_2} e^{-q|x_1-x_2|} dx_1dx_2 \\ =\int_0^a \int_0^{x_1} e^{ik_1x_1} e^{ik_2x_2} e^{-q(x_1-x_2)} dx_2dx_1+\int_0^a \int_{x_1}^a e^{ik_1x_1} e^{ik_2x_2} e^{-q(x_2-x_1)} dx_2dx_1\overset{def}=I_1+I_2. \end{gather}$$ Integrals $I_1$ and $I_2$ can be rewritten as $$\begin{gather} I_1=\int_0^a e^{(-q+ik_1)x_1}\int_0^{x_1} e^{(q+ik_2)x_2} dx_2dx_1, \\ I_2=\int_0^a e^{(q+ik_1)x_1}\int_{x_1}^a e^{(-q+ik_2)x_2} dx_2dx_1. \end{gather}$$

M. Strochyk
  • 8,397
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Divide the inner ($dx_1$) integral into one from $0$ to $x_2$ and one from $x_2$ to $a$. Then you get rid of the modulus $|\cdot|$, and the remaining task is elementary.

user37238
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Joachim W
  • 171