I'm having trouble trying to calculate the following limit. I know the answer is not 0, but after several attempts I am stuck on reducing it.

We have $z_0$ as a constant complex number and a fixed real number $c>0$. For positive real x, let $\alpha = z_0 + x$ and $\alpha' = z_0 - x$ and define the path $\sigma(t) = z_0+itc$ for $t\in[-1,1]. $Find the following limit: $\lim_{x\rightarrow 0}{\int_\sigma \frac{1}{z-\alpha}-\frac{1}{z-\alpha'}dz}$

I know I cannot integrate, and after substituting in the path to the integral, I do not get anything useful. Any hints?

What I did is $\lim_{x\rightarrow 0}{ci\int_{-1}^{1} \frac{1}{itc - x}-\frac{1}{itc + x}dt}$ = $\lim_{x\rightarrow 0}{ci\int_{-1}^{1} \frac{2x}{-c^2t^2 - x^2}dt}$

From here I get stuck since I can pull out the 2x, which would go to 0.

  • $\begingroup$ I think what you're missing is that $z = z_0 + itc$ so $z-\alpha = itc-x$ and $z-\alpha' = itc+x$. $\endgroup$ – Cameron Williams Jan 28 '16 at 2:18
  • $\begingroup$ I used that when parametrizing. It doesn't seem to help though $\endgroup$ – Stefan G. Jan 28 '16 at 2:22
  • $\begingroup$ Woops! Didn't notice a handy trig sub. That seems to do it $\endgroup$ – Stefan G. Jan 28 '16 at 2:53
  • $\begingroup$ Yep :) That's all that was missing. With their choices of variables, it seems like they are trying to do some faux special relativity or something. $\endgroup$ – Cameron Williams Jan 28 '16 at 3:13

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