Does $f'$ analytic imply $f$ analytic?

Question

If $f'$ is known to be analytic, does it mean that $f$ is analytic as well?

I've tried to expand $f$ and then to replace the tail of it by the expansion of $f'$, yet the factorials don't add up. I also tried to start with the known-to-converge expansion of $f'$ yet it was unclear how to move to $f$ (I didn't have integration yet).

If the statement isn't true then how does one prove, for example, that $f(x)=-\log\cos(x)$ is analytic in zero by using the fact that its derivative $\tan(x) = \sum_{n=1}^\infty (-1)^{n-1} 2^{2n}(2^{2n}-1) B_{2n}x^{2n-1}/(2n)!$ is analytic in zero?

Answer 1

• If $g(x):=f'(x)=\sum_n a_nx^n$, then find $b_n$ ($n\ge 1$) such that for $G(x):=\sum_n b_nx^n$ we have $G'=g$.

Hint: $(x^n)'=nx^{n-1}$.

Also calculate the convergence radius.

• If you know that $h'=0\implies h=$const., then you get $b_0$ as $(f-G)'=g-g=0$

Answer 2

since $f'$ is analytic in $\mathbb{c}$, it has a power series expansion say $f'(z)=\displaystyle\sum_{n=1}^{\infty}a_n z^n$ so $\int f'(z)dz= \int\displaystyle\sum_{n=1}^{\infty}a_n z^n$. This implies that $f(z)=\displaystyle\sum_{n=1}^{\infty}\int a_n z^n =\displaystyle\sum_{n=1}^{\infty}\frac{a_n}{n+1} z^{n+1}= \displaystyle \sum_{n=1}^{\infty}b_n z^n$ which is anlytic.