How might one evaluate a contour integral like

In summary, the conversation discusses the evaluation of an integral equation involving a simple pole at z = i*k and the possibility of evaluating it using residues. The conversation also considers the limit as k approaches 0 and concludes that the limit would equal -pi. The final conclusion is that the integral cannot be evaluated.
  • #1
Edwin
162
0
How might one evaluate an integral equation like the following:

I = lim k-> 0+ {ClosedContourIntegral around y [1/(z^2 + k^2)]}, where the contour y is a simple, closed, and positively oriented curve that encloses the simple pole at z = i*K?

Is it possible to evaluate integrals of this form?

Inquisitively,

Edwin
 
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  • #2
Why don't you just evaluate the integral inside the limit using residues?
 
  • #3
\oint

I thought that at first, but since the contour y encloses the simple pole at z=ik and presummably not the other simple pole at z=-ik, what is going to happen when the pole becomes order 2 at k=0? But we're not caring what happens at k=0, just what happens near k=0, so take the curve y as, say, [tex]y: \left| z-ik\right|=k[/tex] so that the other pole at z=-ik is excluded for all k, then by the residue Thm. we have

[tex]\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \mbox{Res}_{z=ik} \left( \frac{1}{z^2 + k^2}\right) [/tex]

but we have

[tex]\frac{1}{z^2 + k^2} = \frac{i}{2k(z + ik)}-\frac{i}{2k(z -i k)}[/tex]

so it turns out that

[tex]\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \left( \frac{i}{2k}\right) = -\frac{\pi}{k} [/tex]
 
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  • #4
So the following would be true?


[tex]{k} \oint_y \frac{dz}{z^2 + k^2} = -pi[/tex]

If you took the limit as k approaches 0 from the right of the expression above, would it still equal -pi?
 
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  • #5
the limit would still be [itex]-\pi[/itex].
 
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  • #6
Perfect! Thanks for the information everyone!

Best Regards,

Edwin
 
  • #7
benorin said:
[tex]\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \left( \frac{i}{2k}\right) = -\frac{\pi}{k} [/tex]

Hello benorin. Looks to me this should be:

[tex]
\oint_y\frac{1}{z^2+k^2}dz=\frac{\pi}{k}[/tex]


Since:

[tex]
\mathop\text{Res}\limits_{z=ik}\frac{1}{z^2+k^2}=-\frac{i}{2k}[/tex]

The answer to the original question then would be:

[tex]
\lim_{k\to 0}\int_{\Lambda(k)}\frac{1}{z^2+k^2}dz=\infty[/tex]

where [itex]\Lambda(k)[/itex] is the contour defined above.

Think so anyway. I'm taking Complex Analysis this semester so . . . may need some help.:confused:
 

1. What is a contour integral?

A contour integral is a type of line integral that is used to evaluate the integral of a complex function along a path in the complex plane. It is represented by the symbol ∮ and is used to find the area under a curve in the complex plane.

2. How is a contour integral evaluated?

A contour integral is evaluated by breaking down the path into smaller segments and then using the fundamental theorem of calculus to calculate the integral along each segment. The individual results are then summed to find the total value of the contour integral.

3. What is the role of the Cauchy-Goursat theorem in evaluating a contour integral?

The Cauchy-Goursat theorem is a fundamental theorem in complex analysis that states that if a function is analytic within a closed contour, then the contour integral of the function is equal to zero. This theorem is used to simplify the evaluation of contour integrals by reducing them to integrals over simpler paths.

4. Are there any special techniques for evaluating contour integrals?

Yes, there are several techniques that can be used to evaluate contour integrals, such as the residue theorem, Cauchy's integral formula, and the method of partial fractions. These techniques are especially useful when dealing with complex functions that have singularities or poles.

5. What are some applications of contour integrals?

Contour integrals have a wide range of applications in mathematics, physics, and engineering. They are used to calculate complex line integrals, solve differential equations, and evaluate path integrals in quantum mechanics. They also have applications in fluid dynamics, electromagnetism, and signal processing.

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