What happens when $\beta_1 + \beta_2=1$ and when $0<\beta_1 + \beta_2<1$? I have the following example of the Scwarz-Christoffel integral formula:
$$S(z)=\int_0^z w^{-\beta_1}(1-w)^{-\beta_2}dw$$
with $0<β_1 <1, 0<β_2 <1$, and $1<β_1 +β_2 <2$ and I know that by the following theorem:

Theorem: Suppose S(z) is the Schwarz-Christoffel formula (the general one). 
(i) If  $\sum_{k=1}^n β_k = 2$, and $P$ denotes the polygon whose
  vertices are given (in order) by $a_1,...,a_n$ , then $S$ maps the
  real axis onto $P−{a_∞}$.  The point $a_∞$ lies on the segment
  $[a_n,a_1]$ and is the image of the point at infinity. Moreover, the
  (interior) angle at the vertex $a_k$ is $α_kπ$ where $α_k = 1 − β_k$. 
(ii) There is a similar conclusion when $1 < \sum_{k=1}^n β_k <2$,
  except now the image of the extended line is the polygon of $n + 1$
  sides with vertices $a_1, a_2, . . . , a_n, a_∞$. The ang le at the
  vertex $a_∞$ is $α_∞π$ with $α_∞=1-\beta_∞$ where
  $\beta_∞=2-\sum_{k=1}^n β_k$

That the above example maps $H$ (upper half plane) to a triangle whose vertices are the images of $0, 1$, and $∞$, and with angles $α_1π, α_2π$, and $α_3π$, where $α_j + β_j = 1$ and $β_1 + β_2 + β_3 = 2$, Am I right?.
The thing is that the above result doesn't take care of the the situation $\beta_1 + \beta_2=1$ and when $0<\beta_1 + \beta_2<1$ in my above example, So Can someone help me with these issues please?
I think I can proceed as follows  

The thing is that  is not clear for my specific function how to manipulate because for example we have the following 
$$S(1)=\int_0^1 w^{-\beta_1}(1-w)^{-\beta_2}dw$$
That it's like a beta function but I don't know which change of variables or how to transform that in a truly beta function and when $z$ is not $1$ how can I apply the method I have said above ? 
Thanks in advance. 

When $\beta_1+\beta_2<1$ then the upper half plane is mapped onto an infinite truncated sector (a sector with the tip cut off). The apex angle of the sector is $(1-\beta_1-\beta_2)\pi$. When $\beta_1+\beta_2=1$ then the upper half plane is mapped onto half a strip with a slanted edge (an infinite strip cut in two at an angle).
