MHB The Residue Theorem To Evaluate Integrals

Click For Summary
The discussion focuses on using the residue theorem to evaluate integrals, highlighting the need for summation of residues and the importance of selecting the correct branch for multi-valued functions like square roots. Participants emphasize the necessity of choosing an appropriate contour, such as a keyhole contour, to apply the theorem effectively. The calculation involves computing four integrals and equating them to 2(pi)i times the sum of the residues. Additionally, it's noted that certain parts of the contour integral can be shown to be negligible. The final evaluation of the integral is given as π/(2√2)√(√5 - 1).
joypav
Messages
149
Reaction score
0
View attachment 7330

I have no idea if this is in the right direction. I know I am going to need the summation of the residues to use the theorem. I found the residues using the limit, but do I need to change these using the euler formula?

We are supposed to be working problems at home and I am getting a bit lost as the semester goes on. I would really appreciate some help or a push in the right direction!
 

Attachments

  • 9ve7wm.jpg
    9ve7wm.jpg
    15.1 KB · Views: 188
Physics news on Phys.org
Square roots are multi-valued, so you have to choose a branch $\log z$ first. Also, you need to choose a contour for which the residue theorem can be applied. Consider using a keyhole contour.
 
Okay, yeah. Something more like this? Lacking details of course, but is this the idea?
View attachment 7336
Then I will have four integrals to compute, set equal to 2(pi)i*sum of the residues.
 

Attachments

  • drbml4.jpg
    drbml4.jpg
    26.4 KB · Views: 134
It looks right so far. Now show that the integral of $f$ becomes negligible over certain parts of the contour.
 
Thanks!
 
For completion, the integral evaluates to

$$\boxed{\frac{\pi}{2\sqrt{2}}\sqrt{\sqrt{5}-1}}$$
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K