1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hi hypernova. Thanks for your answer! Do you mind to have a live chat discussion? I still cannot understand some of the steps above. $\endgroup$ – BeeTiau Apr 27 '18 at 16:12
  • $\begingroup$ Do you mind to expand the derivation from step #2 to #3 and from #3 to #4 above? $\endgroup$ – BeeTiau Apr 27 '18 at 16:14
  • $\begingroup$ Or perhaps, explain what was actually used to derive step #2 to #3 and from #3 to #4 above? $\endgroup$ – BeeTiau Apr 27 '18 at 16:20
  • $\begingroup$ @BeeTiau: Sure! Apology for this delayed response. Can we start a live chat now? Plus, were you asking how the integrand is expanded to series or how the $c_0$ and $c_1$ are determined? $\endgroup$ – hypernova Apr 27 '18 at 16:23
  • $\begingroup$ Hi hypernova, I think if you could help me to expand the derivation, that would great. I think that's better for the benefit of everyone than just to have a live chat that I previously asked for. $\endgroup$ – BeeTiau Apr 27 '18 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.