How Do You Solve This Complex Vector Integral?

  • Context: Graduate 
  • Thread starter Thread starter hiyok
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The forum discussion centers on solving the complex vector integral \(\int d\vec{q} e^{i\vec{q}\vec{r}} \cos(2\theta)\), where \(\theta\) is the angle of vector \(\vec{q}\) and \(\vec{r}\) is an arbitrary vector within a circular domain of radius \(D\). A participant suggested using the identity \(\cos(2\theta) = \frac{e^{2i\theta} + e^{-2i\theta}}{2}\) to simplify the integral. Ultimately, it was concluded that the integral evaluates to zero due to symmetry, although discrepancies arose when comparing results with Wolfram Alpha, which indicated a non-zero outcome.

PREREQUISITES
  • Understanding of vector calculus and integrals
  • Familiarity with complex exponentials and trigonometric identities
  • Knowledge of symmetry properties in integrals
  • Experience with computational tools like Wolfram Alpha
NEXT STEPS
  • Study vector calculus techniques for evaluating integrals over circular domains
  • Learn about the properties of complex exponentials in integrals
  • Explore symmetry arguments in multi-variable integrals
  • Investigate the limitations of computational tools like Wolfram Alpha for complex integrals
USEFUL FOR

Mathematicians, physicists, and students engaged in advanced calculus, particularly those working with vector integrals and complex analysis.

hiyok
Messages
106
Reaction score
0
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
 
Physics news on Phys.org
hiyok said:
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
 
Thanks for your response.

I have worked it out. It amounts to zero by symmetry.
 
amind said:
> 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.
 
eigenperson said:
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.
:(
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K