Residue theorem problem involving real value Intergral

Click For Summary
SUMMARY

The discussion focuses on evaluating the improper integral ∫ from infinity to 0 of x²/(x²+1)(x²+16) using the residue theorem. The relevant singularities are identified at i and 4i, both of which are simple poles. The residues at these poles are calculated as -2i/15 and i/30, respectively. The final result of the integral is determined to be π/10 after multiplying the sum of the residues by 2πi and considering the limits of integration.

PREREQUISITES
  • Understanding of complex analysis, specifically the residue theorem.
  • Familiarity with evaluating improper integrals.
  • Knowledge of singularities and poles in complex functions.
  • Experience with calculating residues for complex functions.
NEXT STEPS
  • Study the residue theorem in detail, focusing on its applications in evaluating integrals.
  • Learn about singularities and their classifications in complex analysis.
  • Explore techniques for calculating residues, including the use of Laurent series.
  • Investigate other examples of improper integrals evaluated using complex analysis methods.
USEFUL FOR

Students and professionals in mathematics, particularly those studying complex analysis, as well as anyone involved in advanced calculus or integral evaluation techniques.

xdrgnh
Messages
415
Reaction score
0

Homework Statement



∫ from infinity to 0 of x^2/(x^2+1)(x^2+16)

Homework Equations



The Integral will be the sum of the residues times 2πi

The Attempt at a Solution



The only singularities that matter are at i and 4i and they are simple poles. So x^2/(x^2+1)/(2x) and I plug in 4i and get -2i/(15) then I plug in i into x^2/(x^2+16)/(2x) and get i/30 and when I add them I get -3i/30. Because it's only from 0 and to negative infinity I mult that by ∏i and I get ∏/10.
 
Physics news on Phys.org
Looks good.

You glossed over several steps, which I'm assuming you understand. Just make sure you know why you can do what you did.
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K