Schwartz-Kristoffel conformal mapping I have the following function from the Schwartz-Kristoffel conformal mapping,
\begin{align}
\begin{split}
\omega(\zeta) &= c\int_{}^{\zeta}\left(1-\dfrac{a_1}{t}\right)^{a_1-1}\left(1-\dfrac{a_2}{t}\right)^{a_2-1}...\left(1-\dfrac{a_k}{t}\right)^{a_k-1}dt \\ &=c\int_{}^{\zeta} \prod_{k=1}^{n-1} \left(1-\dfrac{a_k}{t}\right) ^{\alpha_k-1} dt
\end{split}
\end{align}
It says that since $|a_n|=1$, $|t|>1$, expanding the integrand of the above equation into series and integrating, one obtain,
\begin{align}
\begin{split}
\omega(\zeta)=c\left\lbrace\zeta-[(a_1-1)a_1+(a_2-1)a_2+...+(a_k-1)a_k]\ln\zeta+\dfrac{e_1}{\zeta}+\dfrac{e_2}{\zeta^2}+... \right\rbrace 
\end{split}
\end{align}
Can someone explain how do we get to that last equation above?
 A: Let us make some preparations at first.
Preparation 1. Thanks to the fact that $\left|a_k\right|=1$ and that $\left|t\right|>1$, we have $\left|a_k/t\right|<1$, for which the logarithmic function yields the following expansion
$$
\log\left(1-\frac{a_k}{t}\right)=-\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m.
$$
Preparation 2. Note that the exponential function is holomorphic on the whole complex plane, i.e., $\forall\,z\in\mathbb{C}$,
$$
\exp z=\sum_{p=0}^{\infty}\frac{1}{p!}z^p.
$$
With the above two preparations, we may rewrite the original integrand as follows.
\begin{align}
I&=\prod_{k=1}^{n-1}\left(1-\frac{a_k}{t}\right)^{a_k-1}\\
&=\prod_{k=1}^{n-1}\exp\left[\left(a_k-1\right)\log\left(1-\frac{a_k}{t}\right)\right]&\text{(using $a^b=e^{b\log a}$)}\\
&=\exp\left[\sum_{k=1}^{n-1}\left(a_k-1\right)\log\left(1-\frac{a_k}{t}\right)\right]&\text{(using $e^{a_1}e^{a_2}\cdots=e^{a_1+a_2+\cdots}$)}\\
&=\exp\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]&\text{(using preparation 1)}\\
&=\sum_{p=0}^{\infty}\frac{1}{p!}\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]^p.&\text{(using preparation 2)}
\end{align}
This last line implies that the integrand is no more than a power series with respect to $1/t$. That is, as one expands all brackets and parentheses therein, the result reads
$$
I=c_0+c_1\left(\frac{1}{t}\right)+c_2\left(\frac{1}{t}\right)^2+\cdots=c_0+\frac{c_1}{t}+\frac{c_2}{t^2}+\cdots,
$$
where each $c_j$ is a constant. These constants can be determined from the expression of $I$ from above. For example, $c_0$ is the constant coefficient, which corresponds to the $p=0$ term in the expression (because any $p>0$ term would include $1/t$). When $p=0$, we have
$$
\frac{1}{p!}\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]^p=\frac{1}{0!}\left[-\sum_{k=1}^{n-1}\left(a_k-1\right)\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{a_k}{t}\right)^m\right]^0=1.
$$
Thus
$$
c_0=1.
$$
Likewise, $c_1$ corresponds to the $p=1,m=1$ term, because the least order of $t$ in any $p>1$ term is $\left(1/t\right)^p$, while the least order of $t$ in any $p=1,m>1$ term is $\left(1/t\right)^m$. Taking $p=m=1$, and we eventually have
$$
c_1=-\sum_{k=1}^{n-1}\left(1-a_k\right)a_k.
$$
Now, the integration of the integrand is of the form
$$
\int^{\zeta}I{\rm d}t=\int^{\zeta}\left(c_0+\frac{c_1}{t}+\frac{c_2}{t^2}+\frac{c_3}{t^3}+\cdots\right){\rm d}t=c_0\zeta+c_1\log\zeta-\frac{c_2}{\zeta}-\frac{c_3}{2\zeta^2}+\cdots,
$$
or, by using the above values for $c_0$ and $c_1$,
$$
\int^{\zeta}I{\rm d}t=\zeta-\left(\sum_{k=1}^{n-1}\left(1-a_k\right)a_k\right)\log\zeta+\frac{e_1}{\zeta}+\frac{e_2}{\zeta^2}+\cdots,
$$
where $e_j$'s are constants made up of $c_j$'s, e.g., $e_1=-c_2$ and $e_2=-c_3/2$.
Finally, there is a common factor $c$ in front of all terms. Taking this term into consideration, and one obtains
$$
\omega(\zeta)=c\int^{\zeta}I{\rm d}t=c\left[\zeta-\left(\sum_{k=1}^{n-1}\left(1-a_k\right)a_k\right)\log\zeta+\frac{e_1}{\zeta}+\frac{e_2}{\zeta^2}+\cdots\right],
$$
as is expected.
