Is this integration correct?

1. May 21, 2015

Xyius

I have come across the following integral in my PhD research.

$$\int_{-L_y/2}^{L_y/2}\int_{-L_x/2}^{L_x/2}e^{2 i k \sqrt{(d \cos\theta + x)^2+(d \sin\theta + y)^2}}dxdy$$

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,
$$\int \int e^{2 i k \sqrt{u^2+v^2}}dudv.$$

But then I can make yet another variable substitution by letting,

$r= \sqrt{u^2+v^2} \rightarrow dudv=rdr.$

Which gives me,

$$\int e^{2 i k r}r dr$$

I can easily integrate this by parts, and (assuming I did all my steps correctly) I get the following.

$$\frac{1}{2k}e^{2 i k r}\left[ \frac{1}{2k}-r \right]$$

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?

$r_{min}=\sqrt{(d\cos\theta - L_x/2)^2+(d\sin\theta-L_y/2)^2}$
$r_{max}=\sqrt{(d\cos\theta + L_x/2)^2+(d\sin\theta+L_y/2)^2}$

2. May 21, 2015

RUber

Your method looks good. I didn't check the integration by parts, but I don't doubt that it is correct.
However, when you go to put the limits of integration, it is best to consider the limits with each change of variable.
$u = d\cos\theta + x$ so your limits on $u$ are simply $d\cos\theta-L_x/2 , d\cos\theta+L_x/2$ similarly for $v$.
When you change to $r$, you have to look at the actual minimum value of the variable over the range of u and v.
If $L_x/2 > d\cos \theta \text{ and } L_y/2 > d\sin \theta$ then your minimum will be 0. Consider various other cases.
Your maximum will be at $d \cos \theta + (sign [d\cos\theta]) L_x/2$ and similarly for the y component.
Know the signs of your constants and the domain of theta to make this process simpler.

3. May 21, 2015

RUber

$\int e^{2ikr} r dr$
$\frac{1}{2k}e^{2 i k r}\left[ \frac{1}{2k}-r \right]$

When I checked the integration by parts, I still had some imaginary coefficients, like $\frac{1}{2ik}$.

4. May 21, 2015

Xyius

Thank you for your response! That makes perfect sense. I will go through it and see if it is straight forward or not.

Also yes you are correct, I accidentally dropped a factor of $i$.

5. May 25, 2015

theodoros.mihos

$$dx\,dy = r\,dr\,d\phi$$