Representing power series as a function - what to do with the constant after integration? 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?
 A: 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 $C=0$.
A: You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!
For $|t|<1,$ we have that $$\sum_{n=0}^\infty t^n=\frac1{1-t},$$ so $$\sum_{n=1}^\infty t^n=\frac1{1-t}-1=\frac{t}{1-t}.$$ Thus, we have for $|x|<1$ that $$f'(x)=\sum_{n=1}^\infty x^{3n-1}=\frac1x\sum_{n=1}\left(x^3\right)^n=\frac1x\cdot\frac{x^3}{1-x^3}=\left(1-x^3\right)^{-1}\cdot-\frac13\frac{d\left(1-x^3\right)}{dx}.$$ This has the antiderivative family $$f(x)=-\frac13\ln\left(1-x^3\right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.

More generally, let's suppose that you've used a power series $$\sum_{n=0}^\infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+\sum_{n=0}^\infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be. Namely, we will have $C=f(a)-a_0.$
