Integral Calculation with Complex Analysis - Can Residue Theorem Help?

Click For Summary

Discussion Overview

The discussion revolves around the application of the residue theorem in calculating an integral that involves a singularity at the point 0, which lies on the curve being considered. Participants explore the implications of this singularity and the use of Cauchy Principal value in the context of complex analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant questions the applicability of the residue theorem when the singularity at 0 is on the curve rather than enclosed by it.
  • Another participant suggests using the Cauchy Principal value to address the singularity at 0.
  • A different participant notes that the residue at x=0 is zero, describing it as a removable singularity, and proposes adding an odd function to resolve this issue.
  • Participants discuss the contour integral approach, describing the components of the contour formed by straight lines and semicircles around the singularity.
  • There is a request for clarification regarding the parameters R and r used in the contour description.
  • Further elaboration on the limits of R approaching infinity and r approaching zero is provided, along with the implications for the integral calculations.
  • Participants confirm the understanding of the contour and express a desire to continue the discussion if further difficulties arise.

Areas of Agreement / Disagreement

Participants express differing views on how to handle the singularity at 0 and the application of the residue theorem. There is no consensus on a definitive approach, and the discussion remains exploratory with multiple perspectives presented.

Contextual Notes

Limitations include the dependence on the definitions of the contour and the singularity, as well as the unresolved mathematical steps regarding the integral calculations.

asi123
Messages
254
Reaction score
0
Hey guys.
I need to calculate this integral so I was thinking about using the residue theorem.
The thing is that the point 0 is not enclosed within the curve that I'm about to build, it's on it.
Can I still use the theorem?

Thanks a lot.
 

Attachments

  • 1.jpg
    1.jpg
    1.5 KB · Views: 402
Physics news on Phys.org
So use Cauchy Principal value to bypass the singularity at 0.
 
You have a bigger problem then that, the residue at x=0 is zero (removeable singularity). This is a standard example and is often done as follows
consider
To fix the residue=0 problem add in an odd function the usual one is to use
exp(x i)/x=cos(x)/x+i sin(x)/x
the cos(x)/x diverges as an improper integral, but as lqg states cauch principle sense may be used
now consider ther contour formed by theese pieces
(-R,r) straight line
(-r,r) semicircle arc about z=0
(r, R) straight line
(R,r) semicircle arc about x=0
 
lurflurf said:
You have a bigger problem then that, the residue at x=0 is zero (removeable singularity). This is a standard example and is often done as follows
consider
To fix the residue=0 problem add in an odd function the usual one is to use
exp(x i)/x=cos(x)/x+i sin(x)/x
the cos(x)/x diverges as an improper integral, but as lqg states cauch principle sense may be used
now consider ther contour formed by theese pieces
(-R,r) straight line
(-r,r) semicircle arc about z=0
(r, R) straight line
(R,r) semicircle arc about x=0

I didn't quite understand the contour you described there.
What's R and What's r?

Thanks a lot.
 
asi123 said:
I didn't quite understand the contour you described there.
What's R and What's r?

Thanks a lot.

Take limits R-> +infinity,r->+0
If you draw a pictuis an upper semicircle that in the limit is large with a tiny semicircle at the origin then you get for the various integrals

(-R,r) straight line
-infinity+(Integral you want)/2
(-r,r) semicircle arc about z=0
{+,-}[+ if it was upper - if it was lower] pi*i (residue theorem)
(r, R) straight line
-infinity+(integral you want)/2
(R,r) semicircle arc about x=0
0
 
lurflurf said:
Take limits R-> +infinity,r->+0
If you draw a pictuis an upper semicircle that in the limit is large with a tiny semicircle at the origin then you get for the various integrals

(-R,r) straight line
-infinity+(Integral you want)/2
(-r,r) semicircle arc about z=0
{+,-}[+ if it was upper - if it was lower] pi*i (residue theorem)
(r, R) straight line
-infinity+(integral you want)/2
(R,r) semicircle arc about x=0
0

Is it something like that?

Thanks again.
 

Attachments

  • scan0011.jpg
    scan0011.jpg
    9.7 KB · Views: 374
asi123 said:
Is it something like that?

Thanks again.

That is it. I guess that is one reasone why some people do not like old math books with no pictures. Were you able to finish?
 
lurflurf said:
That is it. I guess that is one reasone why some people do not like old math books with no pictures. Were you able to finish?

I'll try, if I'll have some troubles, I'll be back :smile:.
Thanks a lot.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K