How Do You Solve This Complex Vector Integral?

[hiyok]

Hi,

I'm stumbled on the following integral,

$\int d\vec{q} e^{i\vec{q}\vec{r}} \cos(2\theta)$,

where $\theta$ denotes the angle of the vector $\vec{q}$ and the $\vec{r}$ is an arbitrary vector. The integral domain is a circular area of radius $D$.

Could anyone help me out ?

Many thanks.

hiyok

 

[mathman]

Hi,

I'm stumbled on the following integral,

$\int d\vec{q} e^{i\vec{q}\vec{r}} \cos(2\theta)$,

where $\theta$ denotes the angle of the vector $\vec{q}$ and the $\vec{r}$ is an arbitrary vector. The integral domain is a circular area of radius $D$.

Could anyone help me out ?

Many thanks.

hiyok

Try cos(2θ) =(exp(2iθ)+exp(-2iθ))/2

 

[hiyok]

Thanks for your response.

I have worked it out. It amounts to zero by symmetry.

 

[amind]

> I have worked it out. It amounts to zero by symmetry.
Although , I couldn't solve but it is not zero according to wolfram alpha.
see : http://www.wolframalpha.com/input/?i=∫+(e^(i*q*r)+cos(2θ))dq

 

[eigenperson]

> I have worked it out. It amounts to zero by symmetry.
Although , I couldn't solve but it is not zero according to wolfram alpha.
see : http://www.wolframalpha.com/input/?i=∫+(e^(i*q*r)+cos(2θ))dq

If you're going to ignore both the original poster's notation ($\theta$ has a specific meaning, and it's a vector integral) and the domain of integration, you can't expect to just type the integral into Wolfram Alpha and expect to get the right result.

 

[amind]

If you're going to ignore both the original poster's notation ($\theta$ has a specific meaning, and it's a vector integral) and the domain of integration, you can't expect to just type the integral into Wolfram Alpha and expect to get the right result.

I'm sorry for that , I'm not great in calculus and I probably skimmed through the posts.
:(