http://www.maa.org/sites/default/files/pdf/upload_library/2/Kalman-2013.pdf
On page 44, they conclude that $g'(z) = - \displaystyle\frac{\ln(1-z)}{z}$ by saying that it is just $\displaystyle\frac{f(z)}{z}$. So whats happening is that we would get,
$f(z) = \displaystyle\sum_{k=1}^{\infty}\frac{1}{k}z^k$. Then $f'(z) = \displaystyle\sum_{k=1}^{\infty}z^{k-1} = 1 + z + z^2 + ... = \frac{1}{1-z}$ for $|z| < 1$.
Then if we take $\displaystyle\int\frac{1}{1-z}dz$ we get $f(z) = \displaystyle\sum_{k=1}^{\infty}\frac{1}{k}z^k = -\ln(1-z)$. From here you just divide both sides by $z$ and one gets the result.
Here is my question. If you integrate, one is going to get a constant $C$ with it. Now the paper says to just take $f(0)$ which makes $C = 0$. Of course it will make it zero but I don't understand why you can just randomly do that. I always thought the constant would not be there only if one computed a definite integral. Can this be thought of as a definite integral? The constant is just confusing me because the way I set it up, obviously an indefinite integral will have a $+ C$.
The help would be appreciated!
Equation (1) in the article says $$ f(z) = \sum \frac{1}{k} z^k = z + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4} + \ldots $$ This serves as the definition of the function $f(z)$. Notice that $f$ is not used anywhere in the article until it is introduced in equation (1).
Now, a bit of calculus shows $f(z) = -\ln(1 - z) + C$, and you were OK with that--your question is about how to find $C$.
By plugging in $z=0$ into equation (1), we get: $$ f(0) = 0 + \frac{0^2}{2} + \frac{0^3}{3} + \frac{0^4}{4} + \ldots $$ This series sums to $0$, so we have $f(0) = 0$. Thus $$ 0 = f(0) = -\ln(1 - 0) + C = 0 + C = C, $$ so $C = 0$. Therefore $f(z) = -\ln(1-z)$. $\Box$