Complex Analysis Residue Problem

Click For Summary
SUMMARY

The integral from 0 to infinity of 1/(x^3 + 1) can be evaluated using residue theory, specifically by identifying the singular point at (1+sqrt(3)*i)/2. The polynomial (x^3 + 1) has zeros at x = -1, (1+sqrt(3)*i)/2, and (1-sqrt(3)*i)/2, which can be factored as (x + 1)(x - (1+sqrt(3)*i)/2)(x - (1-sqrt(3)*i)/2). This factorization is confirmed as correct, allowing for the application of the residue theorem to solve the integral.

PREREQUISITES
  • Complex analysis fundamentals
  • Residue theorem application
  • Understanding of contour integration
  • Factorization of polynomials in the complex plane
NEXT STEPS
  • Study the residue theorem in detail
  • Learn about contour integration techniques
  • Practice factoring polynomials in the complex plane
  • Explore examples of integrals evaluated using residues
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on complex analysis and integral calculus, will benefit from this discussion.

tylerc1991
Messages
158
Reaction score
0

Homework Statement



The question asks me to find the integral from 0 to infinity of 1/(x^3 + 1), where I have to use the specific contours that they specify. Now I know that I need to use residues (in fact just one here) and the singular point is (1+sqrt(3)*i)/2. Once I can factor the (x^3 + 1) part then I can take the problem from there. Can someone check me to see if I factored this right?

The Attempt at a Solution



(x^3 + 1) has zeros at x = -1, (1+sqrt(3)*i)/2, and (1-sqrt(3)*i)/2, so when I factor this it becomes (x + 1)(x - (1+sqrt(3)*i)/2)(x - (1-sqrt(3)*i)/2). Then I can continue with the residue and solve the problem, but I need to have factored this correctly. Thank you anyone for your help!
 
Physics news on Phys.org
Yes, you factored correctly. You could check by multiplying all the factors back together.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K