Prove $\int_0^\infty \frac{dx}{x^{1/\alpha}(x+1)}$ goes to zero along countour circling branch point. Background: This is part of the countour integration problem where you show that the circle IV goes to zero as $\ep...--prophetes.ai

Prove $\int_0^\infty \frac{dx}{x^{1/\alpha}(x+1)}$ goes to zero along countour circling branch point. Background: This is part of the countour integration problem where you show that the circle IV goes to zero as $\epsilon \to 0$![enter image description here]( The branch cut is x>0, and solving the integration is not the question, It's proving the small circle goes to zero. Circle II goes to zero following the same proof as below, but I don't see how to show circle IV$\to 0$: $$\lim_{\epsilon \to 0}\int_0^{2\pi}\lvert\frac{\epsilon e^{i \theta} i d\theta}{\sqrt[\alpha]{\epsilon} e^{i \theta/\alpha}(\epsilon e^{i \theta}+1)}\rvert\le$$ $$\lim_{\epsilon \to 0}\int_0^{2\pi}\frac{\epsilon d\theta}{\sqrt[\alpha]{\epsilon} (\epsilon-1)}$$ After two applications of l'hopital's rule I get $\frac{\epsilon^{-1/\alpha}(\alpha-1)}{(1+\alpha)}$ which goes to infinity. What did I do wrong? Am I going about this wrong?

The key is to have $$ \lim_{r\to0}\frac{2\pi r}{r^{1/a}}=0 $$ so you need $1-\frac1a\gt0$. That would be $a\lt0$ or $a\gt1$.

The $2\pi r$ is the circumference of the small circle and $r^{-1/a}$ is the absolute value of $\frac1{x^{1/a}}$ on the circle.

The factor of $\frac1{x+1}$ is near $1$ close to the branch point.