The function $f(z)=\frac{1}{(z-2)(z-3)}$ has simple poles at $z=2$ and $z=3$. Is it possible to have a continuum of poles on the real (or imaginary) line (say, from $z=2$ to $z=3$, at all points there is a pole) instead of at discrete values?
Updates
In the comment below by @Michael gave an interesting example $$f(z)=1/({|z|-a}).$$ What kind of singularity is this? Also, what about something like $$f(z)=\int\limits_{-1}^{-1}\frac{da}{z-a}$$ where $a$ is continuous real parameter? What kind of a singularity is this?
Does any of these be called continuum of poles?
For $g$ continuous or $L^1_{loc}$ let $$f(z)=\int_{-r}^r \frac{g(a)}{z-a}da, \qquad m= \frac1{2r}\int_{-r}^r g(a)da, \qquad F(z) = \int_{-r}^r \frac{g(a)-m}{z-a}da$$
Then $F$ is analytic on $\Bbb{C}-[-r,r]$ and $[-r,r]$ is a branch cut and $$\lim_{|z|\to \infty} z F(z) = 0$$
Conversely if $F$ is analytic on $\Bbb{C}-[-r,r]$ and $\lim_{|z|\to \infty}zF(z) = 0$, with $C_R$ a curve enclosing $[-r,r]$ at a distance $1/R$, by the Cauchy integral formula $$F(z) =\lim_{R\to \infty}\frac1{2i\pi}\int_{|z|=R}\frac{F(s)}{s-z}ds -\frac1{2i\pi}\int_{C_R} \frac{F(s)}{s-z}ds= \int_{-r}^r \frac{g(a)}{a-z}da$$
(here $g$ doesn't have to be $L^1$ : it can be a distribution and even an analytic functional..)
If $g$ is analytic then $F(z) = \int_\gamma \frac{g(a)-m}{z-a}da + b$ for any curve $\gamma : -r\to r$ and hence $F$ is analytic on $\Bbb{C}-\{-r,r\}$ with two branch points at $-r,r$.