Why does the following integral starts from $0$? Consider $$f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^{3n+1}$$
It's a power series with a radius, $R=1$. at $x=1$ it converges. Hence, by Abel's thorem:
$$\lim_{x\to 1^-} f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$
Evaluating the derivative
$$f'(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} (3n+1)x^{3n+1} = \ldots = \frac{1}{1+x^3}$$
Now, consider this claim: "Since $f(0) = 0$: $$f(x) = \int_0^x \frac{1}{1+t^3} \ dt$$
I am familiar with the fundamental theorem of calculus, yet not sure why this claim true, More presicely; Why is the integral starts at $0$?
Thanks.

The fundamental theorem tells you that
$$f(x)=f(a)+\int_a^x f'(t) dt.$$
It's convenient to choose $a$ such that $f(a)=0$, because then
$$f(x)=f(a)+\int_a^x f'(t) dt = \int_a^x f'(t) dt.$$
Since your function is a power series with no constant term, it's not hard to see that you can use $a=0$ for this purpose.
