Let $γ\colon[-1,1]\to\mathbb{C}$ , $γ(t)= z_0 + itc$ , $z_0$ fixed and c>0
Prove for x>0 $$\lim_{x\to0} \frac{1}{2πi} \int_γ \left(\frac{1}{z-w} - \frac{1} {z-w'}\right)dz = -1$$
Where $w=z_0 + x$ , $w'= z_0 - x$
I understand that you have to substitute in w and w' but I can't figure out what to do with $$\frac{1}{z-z_0 -x} - \frac{1}{z-z_0 +x}$$ What is the next step? Do I Use the $γ(t)$ function?
Any help on this question would be appreciated. Thanks
On
Making the substitution $z=z_0+itc$ in the second integral gives
$$\int_\gamma\frac{1}{z-z_0+x}dz = \int_{-1}^1\frac{1}{itc+x}ic\,dt = \Bigl[\log(itc+x)\Bigr]_{-1}^1 = \log(x+ic)-\log(x-ic).$$
Using the principal value of $\log$,
$$\log(x \pm ic) = \frac12\log(x^2+c^2) \pm i\tan^{-1}(c/x)$$
so
$$\int_\gamma\frac{1}{z-z_0+x}dz = 2i\tan^{-1}(c/x) \to \pi i \text{ as }x \to 0^+.$$
Some hints:
Write $a>0$ instead of $x$. You may assume $z_0=0$. Then we have to look at the integral $$\int_{-ic}^{ic}\left({1\over z-a}-{1\over z+a}\right)\ dz\ ,$$ where the integration is along the segment $\sigma$ on the imaginary axis connecting $-ic$ and $ic$. Now use the fact that $${dz\over z-a}=d\log\bigl(|z-a|\bigr)+i\ d\arg(z-a)\ .$$ As $|ic\pm a|=|{-ic}\pm a|$ we only have to check how $\arg(z\pm a)$ changes along $\sigma$. This can be described in terms of the angles in the triangle with vertices $-ic$, $ic$, $a$, resp., $-ic$, $ic$, $-a$. (Draw a figure!)
Finally, analyze what happens when $a\to {0+}$ for fixed $c>0$.