- #1
Xyius
- 508
- 4
I have come across the following integral in my PhD research.
[tex]\int_{-L_y/2}^{L_y/2}\int_{-L_x/2}^{L_x/2}e^{2 i k \sqrt{(d \cos\theta + x)^2+(d \sin\theta + y)^2}}dxdy[/tex]
Ultimately, x and y are integrated out and I will be left with a function of ##\theta##. I am doing this the following way. First I make the following substitutions.
##u=d\cos\theta+x \rightarrow du=dx ##
## v=d\sin\theta+y \rightarrow dv=dy ##
So then I get,
[tex]\int \int e^{2 i k \sqrt{u^2+v^2}}dudv.[/tex]
But then I can make yet another variable substitution by letting,
## r= \sqrt{u^2+v^2} \rightarrow dudv=rdr. ##
Which gives me,
[tex]\int e^{2 i k r}r dr[/tex]
I can easily integrate this by parts, and (assuming I did all my steps correctly) I get the following.
[tex]\frac{1}{2k}e^{2 i k r}\left[ \frac{1}{2k}-r \right][/tex]
Now here is where it gets tricky for me
I now need to back substitute for ##r## to get it in terms of my original variables ##x## and ##y##. When it comes to evaluating the limits, can I just use the expression I got (after substituting back) and first evaluate for ##x## and then for ##y##?
Or perhaps I can just leave r in and just use the min and max values of u and v to write the limits of r?
##r_{min}=\sqrt{(d\cos\theta - L_x/2)^2+(d\sin\theta-L_y/2)^2}##
##r_{max}=\sqrt{(d\cos\theta + L_x/2)^2+(d\sin\theta+L_y/2)^2}##
[tex]\int_{-L_y/2}^{L_y/2}\int_{-L_x/2}^{L_x/2}e^{2 i k \sqrt{(d \cos\theta + x)^2+(d \sin\theta + y)^2}}dxdy[/tex]
Ultimately, x and y are integrated out and I will be left with a function of ##\theta##. I am doing this the following way. First I make the following substitutions.
##u=d\cos\theta+x \rightarrow du=dx ##
## v=d\sin\theta+y \rightarrow dv=dy ##
So then I get,
[tex]\int \int e^{2 i k \sqrt{u^2+v^2}}dudv.[/tex]
But then I can make yet another variable substitution by letting,
## r= \sqrt{u^2+v^2} \rightarrow dudv=rdr. ##
Which gives me,
[tex]\int e^{2 i k r}r dr[/tex]
I can easily integrate this by parts, and (assuming I did all my steps correctly) I get the following.
[tex]\frac{1}{2k}e^{2 i k r}\left[ \frac{1}{2k}-r \right][/tex]
Now here is where it gets tricky for me
I now need to back substitute for ##r## to get it in terms of my original variables ##x## and ##y##. When it comes to evaluating the limits, can I just use the expression I got (after substituting back) and first evaluate for ##x## and then for ##y##?
Or perhaps I can just leave r in and just use the min and max values of u and v to write the limits of r?
##r_{min}=\sqrt{(d\cos\theta - L_x/2)^2+(d\sin\theta-L_y/2)^2}##
##r_{max}=\sqrt{(d\cos\theta + L_x/2)^2+(d\sin\theta+L_y/2)^2}##