Cauchy P.V. of an Improper Integral

In summary, it's not clear whether choosing a semicircle CR on the lower half-plane below the real axis is valid or not.
  • #1
Go37Pi
2
0
I was doing a Fourier Transform Integral, and was wondering if it would be legitimate for me to choose a semicircle CR on the lower half-plane below the real axis rather than choosing a semicircle CR on the upper half-plane above the real axis. I would expect it to be valid because the contour integral should have equal areas regardless of whether it is oriented above or below the Real axis. Is this right to expect?
 
Physics news on Phys.org
  • #2
It's not clear whether you're asking about the Fourier transform or its inverse. I presume your interest is in the inverse transform.

You can decide which contour to used based on whether [itex]e^{i \omega t}[/itex] diverges as [itex]\omega[/itex] goes to [itex]i \infty[/itex] or [itex]-i \infty[/itex] and that depends on the sign of t.
 
  • #3
Go37Pi said:
I was doing a Fourier Transform Integral, and was wondering if it would be legitimate for me to choose a semicircle CR on the lower half-plane below the real axis rather than choosing a semicircle CR on the upper half-plane above the real axis. I would expect it to be valid because the contour integral should have equal areas regardless of whether it is oriented above or below the Real axis. Is this right to expect?
You should not think in terms or *area*. You really are doing a *line* integral. So it's not a question of whether the area is the same, it's the question of whether the contribution *along* the upper and lower contours is the same. What you must ensure is that the contribution along the infinite semi-circle vanishes. As Tide pointed out, this is determined by the sign of the exponent in omega t. For example, if you have [itex] e^{-i \omega t} [/itex], if you close with a contour above the real axis, as t goes to [itex] + i \infty [/itex], the exponent will go to [itex] e^{ + \infty} [/itex] which means that the contribution along that contour will blow up! So you *must* close below in that case.
 

Related to Cauchy P.V. of an Improper Integral

1. What is the Cauchy P.V. of an Improper Integral?

The Cauchy Principal Value (Cauchy P.V.) of an improper integral is a way of assigning a finite value to an integral that would otherwise be undefined due to a singularity or infinite discontinuity in the integrand. It is denoted by the symbol P.V. and is calculated using the limit of the integral as the singularity approaches a specific point.

2. How is the Cauchy P.V. different from the regular value of an integral?

The Cauchy P.V. differs from the regular value of an integral because it takes into account the behavior of the integrand near a singularity. The regular value of an integral is calculated by integrating over a specific interval, while the Cauchy P.V. involves taking the limit of the integral as the singularity approaches a certain point.

3. When is the Cauchy P.V. used in mathematical applications?

The Cauchy P.V. is commonly used in mathematical applications involving singularities, such as in the study of complex analysis and Fourier transforms. It is also frequently used in physics, particularly in the calculation of electric and magnetic fields around point charges or moving particles.

4. What is the significance of the Cauchy P.V. in mathematics?

The Cauchy P.V. is significant in mathematics because it allows for the evaluation of integrals that would otherwise be undefined due to singularities. This makes it a valuable tool for solving complex mathematical problems and allows for a better understanding of the behavior of functions near singular points.

5. Are there any limitations to using the Cauchy P.V. method?

While the Cauchy P.V. is a useful method for calculating integrals, it does have some limitations. It can only be used for integrals with finite limits and cannot be applied to certain types of singularities, such as essential singularities. Additionally, it may not always provide a unique solution for the integral, as there can be different paths to approach the singularity that result in different Cauchy P.V. values.

Similar threads

  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • General Math
Replies
2
Views
1K
  • Topology and Analysis
Replies
4
Views
2K
Back
Top