How might one evaluate a contour integral like

[Edwin]

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

 

[shmoe]

Why don't you just evaluate the integral inside the limit using residues?

 

[benorin]

\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, $$y: \left| z-ik\right|=k$$ so that the other pole at z=-ik is excluded for all k, then by the residue Thm. we have

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

but we have

$$\frac{1}{z^2 + k^2} = \frac{i}{2k(z + ik)}-\frac{i}{2k(z -i k)}$$

so it turns out that

$$\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \left( \frac{i}{2k}\right) = -\frac{\pi}{k}$$

 

[Edwin]

So the following would be true?


$${k} \oint_y \frac{dz}{z^2 + k^2} = -pi$$

If you took the limit as k approaches 0 from the right of the expression above, would it still equal -pi?

 

[Data]

the limit would still be $-\pi$.

 

[Edwin]

Perfect! Thanks for the information everyone!

Best Regards,

Edwin

 

[saltydog]

$$\oint_y \frac{dz}{z^2 + k^2} = 2\pi i \left( \frac{i}{2k}\right) = -\frac{\pi}{k}$$

Hello benorin. Looks to me this should be:

$$\oint_y\frac{1}{z^2+k^2}dz=\frac{\pi}{k}$$


Since:

$$\mathop\text{Res}\limits_{z=ik}\frac{1}{z^2+k^2}=-\frac{i}{2k}$$

The answer to the original question then would be:

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

where $\Lambda(k)$ is the contour defined above.

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