0
$\begingroup$

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 $

$\endgroup$

2 Answers 2

2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .