Residues at poles What is the residue of $$f(x)=\frac{1}{(x^2+1)^a}$$ at $x^2=\pm i$, where $0<a<1$ ? My intuition tells me that there must be a non-zero residue, but my attempts to compute tells me the residue is $0$. How can this be so when $x^2+1=0$ when $x=\pm i$ ?
 A: Although $(1+z^2)^a$ is not analytic in a neighborhood of $i$ or $-i$, we can still compute the circular integral around each point missing the branch cut.
$\log(1+z^2)$ can be well-defined in a domain cut so that if a closed path circled $i$, it also circles $-i$. For example, we could have a branch cut that connects $i$ and $-i$ or a branch cut that extends from $i$ to $\infty$ and another cut that extends from $-i$ to $\infty$.
On any such domain we can then define $(1+z^2)^{\large a}$ via the exponential function. In any case, near $i$,
$$
|f(z)|\sim2^{\large-a}|z-i|^{\large-a}
$$
On a small circle of radius $r$, the length of a circular path is $2\pi r$ and the value of the function would be $\sim2^{\large-a}r^{\large-a}$ the integral around the circle would be at most $\sim2^{\large-a}r^{1\large-a}\to0$ if $0\lt a\lt1$.
Thus, even though we cannot form a closed circuit around $i$ because of branch cuts, the integral around the point vanishes as the radius goes to $0$.
Caveat: Although we have a $0$ "residue" at $i$ and $-i$, this cannot be extended to any useful contour. To extend the result for the small circle to a larger path, the paths are usually connected by two superimposed connectors oppositely directed that cancel each other. Here, the connectors would have to follow the branch cut and the function is not continuous across the branch cut so the integrals would not necessarily cancel.
For example, consider $f(z)=z^{1/2}$ with a branch cut along the positive real axis.
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Limiting to the real axis from above, $f(z)\to\sqrt{x}$, the normal positive square root.
Limiting to the real axis from below, $f(z)\to-\sqrt{x}$, the negative square root.
Let's try to use the same construction to show that the integral along two contours that circle the same singularities are the same. The only place to add the connectors, and keep the contour inside the domain of definition of $f$, is on each side of the branch cut. However, the integral along the connectors do not cancel; in fact, they actually reinforce.
The integral counterclockwise around a circle of radius $r$ is
$$
\begin{align}
\int_0^{2\pi}r^{1/2}e^{i\theta/2}\,\mathrm{d}re^{i\theta}
&=\int_0^{2\pi}r^{3/2}ie^{i3\theta/2}\,\mathrm{d}\theta\\
&=\left.\frac23r^{3/2}e^{i3\theta/2}\right]_0^{2\pi}\\
&=-\frac43r^{3/2}
\end{align}
$$
This makes sense. As shown above, the integral along the circle as $r\to0$ is $0$. The integral along each of the connectors is
$$
\int_0^r\sqrt{x}\,\mathrm{d}x=\int_r^0-\sqrt{x}\,\mathrm{d}x=\frac23r^{3/2}
$$
So the total along all the contours is $0$. However, the point is that the integral along the circles is not constant since the integrals along the connectors do not cancel.
