Integration using residues $\int z^2 \log [(z+1)/(z-1)] dz$

$$\int z^2 \log [(z+1)/(z-1)] dz$$ taken around circle $$|z|=2$$

I am taking residues at $$\pm 1$$.

This gives me 0 as the value of integral. Is this correct.

How do I modify the integral to get value over half the circle?

• The residue at $z=\pm 1$ does not make sense. You should take the residue at infinity. – Kemono Chen Dec 26 '18 at 5:46
• I need to take residues at poles right and those poles which lie inside the |z|=2 – Sonal_sqrt Dec 26 '18 at 5:47
• Yes. $\pm1$ are not poles, they are branch points. – Kemono Chen Dec 26 '18 at 5:47
• But Cauchy residue theorem requires that poles be inside |z|=2, which as I see it there are no such poles. Then we can't apply residue theorem can we? – Sonal_sqrt Dec 26 '18 at 5:58

$$\int_C f(z)dz=-2\pi i \operatorname{Res}_{z=\infty}f(z)$$ But we have $$f(z)=2z+\frac23z^{-1}+o(z^{-1})$$ as $$z\to\infty$$, therefore the integral equals $$\frac43 \pi i$$. $$f$$ is not meromorphic in $$|z|<2$$ , so we can not apply residue theorem inside the circle, but we can apply it outside the circle.