Logarithm real analytic on $(0,\infty)$ I am working on a Tao Analysis II question. I have to prove that $log$ the inverse function of $exp$ is real analytic on $(0,\infty)$. I have already  proven that 
$$
 \forall x \in(-1,1): ln(1-x) = - \sum_{n=1}^\infty \frac{x^n}{n}
$$ and that
$$
\forall x \in (0,2): ln(x)= \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (x-1)^n
$$
Does this help ? Further i may not make use of complex numbers.

Suppose that $f'=g$ on an interval $(a,b)$, and $g$ is real analytic.  For each $c\in(a,b)$, there is an $r>0$ and a sequence $(a_n)_n$ such that for all $x\in(c-r,c+r)$, $g(x)=\sum\limits_{n=0}^\infty a_n(x-c)^n$.    The power series $h(x)=\sum\limits_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}$ also converges for $x\in(c-r,c+r)$, and $h'=g$ in this interval (this requires justification).  Therefore $f(x)=f(c)+h(x)=f(c)+\sum\limits_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}$ for all $x\in(c-r,c+r)$ (e.g., this follows from the Mean Value Theorem).  This shows that $f$ is real analytic.

$g(x)=1/x$ is real analytic on $(0,\infty)$, because for each $c\neq 0$, $g(x)=\sum\limits_{n=0}^\infty\frac{(-1)^n}{c^{n+1}}(x-c)^n$ when $|x-c|<|c|$, by the geometric series identity. (More generally, you may have seen that a quotient of real analytic functions is real analytic away from zeros of the denominator.)
