Representing power series as a function – what to do with the constant after integration

Question

This power series $$f(x)=\sum_{n=1}^{\infty} {\frac{x^{3n}}{3n}}$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=\frac{1}{x(1-x^3)} $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=\log(x)-\frac{1}{3}\log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?

Answer 1

Differentiating $f(x)$:

$$f(x)=\frac{x^3}{3}+\frac{x^6}{6}+\cdots\implies f'(x)=x^2+x^5+\cdots=\sum_{n=1}^{\infty}x^{3n-1}=\frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-\frac13\ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.