- #1
Integral Calculation with Complex Analysis - Can Residue Theorem Help?
- Thread starter asi123
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In summary, the conversation discusses the use of the residue theorem in calculating an integral, but the issue of the point 0 not being enclosed within the curve arises. It is suggested to use Cauchy Principal value to bypass the singularity at 0, but another problem arises with the residue at x=0 being zero. To solve this, an odd function is added and a contour is formed by combining straight lines and semicircle arcs. The conversation ends with discussing the need for pictures in math books and the person planning to try and finish the problem.
- #2
MathematicalPhysicist
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So use Cauchy Principal value to bypass the singularity at 0.
- #3
lurflurf
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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
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
- #4
asi123
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lurflurf said:
I didn't quite understand the contour you described there.
What's R and What's r?
Thanks a lot.
- #5
lurflurf
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asi123 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
- #6
- #7
lurflurf
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asi123 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?
- #8
asi123
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lurflurf said:
I'll try, if I'll have some troubles, I'll be back
.Thanks a lot.
FAQ: Integral Calculation with Complex Analysis - Can Residue Theorem Help?
What is the Residue Theorem?
The Residue Theorem is a powerful tool in complex analysis that allows for the calculation of certain types of integrals. It states that the integral of a function around a closed curve is equal to the sum of the residues of the function at its singular points inside the curve.
How does the Residue Theorem help with integral calculation?
The Residue Theorem simplifies the calculation of certain types of integrals by reducing them to a sum of residues, which are much easier to compute. This can save a significant amount of time and effort in solving complex integrals.
Can the Residue Theorem be used for any type of integral?
No, the Residue Theorem is specifically useful for integrals that involve rational functions or functions with simple poles. It cannot be used for integrals with essential singularities or branch points.
Are there any limitations to using the Residue Theorem?
Yes, the Residue Theorem can only be used for integrals that are taken along closed curves. It also requires knowledge of complex analysis and the ability to identify singular points and their corresponding residues.
What are some applications of the Residue Theorem?
The Residue Theorem has many applications in physics, engineering, and mathematics. It can be used to solve problems involving contour integrals, real integrals, and infinite series. It is also used in the study of complex functions, such as in the calculation of residues and poles.