Determine the region in which $g(z) = \int_{0}^{1} (t-z)^{-1} dt$ is analytic. Does it have any isolated singularities? Does it have any singularities that are not isolated?
My solution: $g$ is not defined on $[0, 1]$ because if $z \in [0, 1]$, then for $t = z$, $(t-z)^{-1}$ is not defined. Outside of $[0, 1]$, $g(z) = log (1- \frac{1}{z})$, which is analytic. So $[0, 1]$ are the singularities and they are non - isolated.
Please verify the solution. Thanks.
Your argument is not rigorous enough but the conclusion is correct. We have $\frac {g(z+h)-g(z)} h=\int_0^{1}\frac 1 {(t-z)(t-z-h)}dt \to \int_0^{1}\frac 1 {(t-z)^{2}}dt$. To justify this you need the fact that if $z \notin [0,1]$ then there is a positive distance from $z$ to $[0,1]$ and you can justify interchange of limit and integral using DCT. Conclusion: $g$ is differentiable, hence analytic on $\mathbb C \setminus [0,1]$.