Since the problem of determining the $a_i$ is quite difficult, we can instead try to observe what happens if we know what the $a_i$ are.

1. Let $a_1,\dots ,a_n\in \mathbb{R}$ with $a_1<\cdots <a_n$
2. Choose ${\alpha }_k\in (0,2)$ such that $\sum _{k=1}^n{\alpha }_k=n-2$
   TASK
   Find the image of $\mathbb{H}$ under the map 1;;$f(z)={\int }_0^z(\zeta -a_1{)}^{{\alpha }_1-1}\dots (\zeta -a_n{)}^{{\alpha }_n-1}d\zeta$. We adopt the convention that for each term $(\zeta -a_k{)}^{a_k-1}$, say that $arg(\zeta -a_k)=\{\begin{array}{ll}0& \zeta \in \mathbb{R},\zeta >a_k\\ \pi & \zeta \in \mathbb{R},\zeta <a_k\end{array}$. With this convention, we have that $arg(\zeta -a_k{)}^{{\alpha }_k-1}\in$1;;$(-\pi ,\pi )$, so our argument here is well defined.

With thees conventions, here are some notes on the shape this could give us:

1. First we notice that as $\zeta \to \mathrm{\infty }$, the integrand behaves like ${\zeta }^{{\alpha }_1-1+\cdots +{\alpha }_n-1}={\zeta }^{-2}$. We get an expression like ${\zeta }^{-2}(1+\mathcal{O}\left(\frac{1}{\zeta }\right))$, which insures 1;;that our integral converges absolutely on the real line.
2. Now we consider our integrand on the intervals $(a_k,a_{k+1})$. For $\zeta \in (a_k,a_{k+1})$, we have that $arg(\zeta -a_i{)}^{{\alpha }_i-1}=0$ for $i\le k-1$, and $arg(\zeta -a_k{)}^{{\alpha }_k-1}=\pi$ for $i\ge k$. Therefore we conclude that the 1;;integrand is constant on this open interval, and therefore the image is a straight line.
3. Now it is also important to consider the integrand on $(a_n,\mathrm{\infty })\cup (\mathrm{\infty },a_1)$ (same infinity since they sit on the Riemann Sphere). We can also show that the integrand is constant on these two intervals, so the image is again a line. Since 1;;our integral converges at infinity, we see that these images are lines connecting $f(a_n)$ to $f(\mathrm{\infty })$, and $f(\mathrm{\infty })$ to $f(a_1)$. I think here we can use the property of the angles to conclude that the image really is a line from $f(a_n)$ to $f(a_1)$.
4. Now conditions on our angles give us the following data:
   1. If all ${\alpha }_k\in (0,1)$, then 1;;the image under $f$ is a convex polygon.
   2. If some of the ${\alpha }_k\in (1,2)$, then the image will have some small interior angles, and this may not be convex. In fact, the boundary could 1;;have some self intersections, so it may not be of the form we had before.
5. The case for $a_n=\mathrm{\infty }$ can be interpreted by the case where $\sum {\alpha }_k<n-2$, and we define ${\alpha }_n=n-2-\sum {\alpha }_k$. We then omit the $a_n$ term from the integral, and much of this works similarly.
6. Finally, for small $n$ we can determine $f$.
   1. For $n=3$, all angles are less than $1$, and so the image will always be convex. Moreover, the shape will be a triangle.
   2. For $n=4$, only one angle can be large, so the image is a square or a dented square.
   3. For $n=5$ CHAOS
7. We may also note that for $a_k\in S^1$ in cyclic order, we can adapt these definitions to 1;;maps from $\mathbb{D}$ to a polygon, using just a conformal mapping (möbius).