about the independence of the angle at $\infty$ of the Schwarz-Christoffel integral In Stein's complex analysis, he said about the integral $$S(z)=\int_{0}^{z} \frac{d\zeta}{(\zeta-A_1)^{b_1}\cdot\cdot\cdot(\zeta-A_n)^{b_n}}$$(the Schwarz-Christoffel integral) where $A_1<A_2<\cdot\cdot\cdot<A_n$ are n distince points on the real axis arranged in increasing order. And if $\Sigma_{k=1}^{n}b_k>1$, then Cauchy's theorem implies $\lim_{r\to\infty}S(re^{i\theta})$ exists and is independent of the angle $\theta$, $0<\theta<\pi$.
Can anyone tell me why this integral is independent of $\theta$？ He seems to have used Cauchy's theorem on a simply connected set which is any integral of a same holomorphic function on the set from 2 points would be the same regardless of the path between the two points. But I don't understand how it is used here. Thanks

You can compare two different angles $\theta_1,\theta_2$ in the following way:  Cauchy's theorem shows that
$S(re^{i\theta_2}) = S(re^{i\theta_1}) + \int_\gamma f(z) dz$
where $f(z)$ is the Schwarz-Christoffel integrand and $\gamma$ is a semi-circular arc of radius $r$ from $re^{i\theta_1}$ to $re^{i\theta_2}$. If you now do the usual arclength integral estimate you can show as $r$ tends to $\infty$, the two values become arbitrarily close.
