I have encountered several times where I need to calculate a contour integral around a circle or half circle with an infinitesimal radius that is enclosing a branch point. In the cases I have encountered so far, the integral converged to zero (I don't suppose that's always the case). But usually, my book skips the justification and leaves it as obvious why it's zero (Maybe it is obvious, but I can't see it). So I was hoping you could help me showing why it's true.
Consider for example the following case:
define $$
f(z)=\frac{1}{\left(1+z^{2}\right)\left(1-z^{2}\right)^{1 / 2}}
$$
I want to integrate it around a half-circle above the branch point at $z=1$.
Making the parameterization $z=1+re^{i\theta}$, $dz=ire^{i\theta}d\theta$ and integrating from $\pi$ to $0$:
$$
\lim _{r \rightarrow 0^{+}} \int_{\pi}^{0} \frac{1}{\left(1+\left(1+r e^{i \theta}\right)^{2}\right)\left(1-\left(1+r e^{i \theta}\right)^{2}\right)^{1 / 2}} i r e^{i \theta} d \theta
$$
Which supposedly gives zero. But I don't think it's an easy integral to solve by hand, and I can only interchange the limit and the integral if it's uniformly convergent, which I believe is beyond the scope to prove. Since the branch point isn't a simple pole, I can't evaluate it to be half the residue. So where do I go from here?
I tried playing with some inequalities taking the absolute value of the integrand, but I couldn't find an expression that proved it converged to zero. $$
\lim _{r \rightarrow 0^{+}} \int_{\pi}^{0} \frac{1}{\left(1+\left(1+r e^{i \theta}\right)^{2}\right)\left(1-\left(1+r e^{i \theta}\right)^{2}\right)^{1 / 2}} i r e^{i \theta} d \theta \leq \lim _{r \rightarrow 0^{+}} \int_{\pi}^{0} \frac{r}{\left| \left(1+\left(1+r e^{i \theta}\right)^{2}\right)\left(1-\left(1+r e^{i \theta}\right)^{2}\right)^{1 / 2}\right|} d \theta
$$ I have trouble finding an appropiate denominator, where the theta dependence dissapear. Using the reverse triangle inequality on the denominator doesn't give me an appropiate expression.
Best Answer
For $r < 1/2$, say, we have that $\lvert1 + r e^{i \theta}\rvert > \frac{1}{2}$, and so $1/\lvert (1 + (1 + re^{i \theta})^2) \rvert < 4/3$ is absolutely bounded above by a constant independent of $\theta$. As $$(1 - (1 + re^{i \theta})^2) = r(2 e^{i \theta} + r e^{2 i \theta}),$$ for $r < 1/2$ again (say), we have that $1/\lvert (1 - (1 + re^{i \theta})^2) < \frac{1}{r} \frac{1}{\lvert 2 e^{i \theta} + re^{i \theta} \rvert} < \frac{1}{r}$. Together, this means that the integrand in $$\int_{\pi}^{0} \frac{1}{\left(1+\left(1+r e^{i \theta}\right)^{2}\right)\left(1-\left(1+r e^{i \theta}\right)^{2}\right)^{1 / 2}} i r e^{i \theta} d \theta$$ is bounded in absolute value by $\frac{4}{3}\sqrt{r}$. Thus the integral is bounded by $\frac{4 \pi}{3} \sqrt{r}$, and goes to $0$ as $r \to 0$.
From a slightly longer perspective, the point is that behavior in $r$ in the integrand looks like $r / (1 \cdot \sqrt{r}) = \sqrt{r}$.