How do I show $\arctan x$ is real analytic on $\mathbb{R}$?
First of all we have $$ \arctan x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \, \forall\, x\in[-1,1] $$ because $$ \arctan x = \int_0^x \frac{dt}{1+t^2} = \sum_{n=0}^{m-1} (-1)^n \frac{x^{2n+1}}{2n+1} +(-1)^{m}\int_0^x\frac{t^{2m}}{1+t^2}\,dt $$ where for all $x\in[-1,1]$ $$ \int_0^x\frac{t^{2m}}{1+t^2}\,dt \le \int_0^x{t^{2m}}\,dt =\frac{|x|^{2m+1}}{2m+1}\to 0 \,(n\to\infty) $$
Then to prove that $\arctan x$ is real analytic on $\mathbb{R}$ we certainly should consider the Taylor expansion at other points because $\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$ doesn't converge when $|x|>1$.
My attempt
I wanted to avoid evaluating $\arctan^{(n)} a$ but work it out via using the property of power series.
Because we have $$ \arctan x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \, \forall\, x\in[-1,1] $$ and $$ \mathrm{arctanh}\, x = \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} \, \forall\, x\in(-1,1)$$ thus $\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$ is absolutely convegent for $x\in(-1,1)$. We can rearrange it as $$ \arctan x = \sum_{n=0}^\infty a_n (x-x_0)^n \,\, \forall\, x\in(-1,1) $$ where $$a_n=\frac{\arctan^{(n)} x_0}{n!}$$
Denote $M=\max\{|-1-x_0|,|1-x_0|\}$.
Thus we have $$ \arctan (x+x_0)=\sum_{n=0}^\infty a_n x^n \,\, \forall\, x\in(-1-x_0,1-x_0) $$ which leads to $$ \arctan (x+x_0)=\sum_{n=0}^\infty a_n x^n \,\, \forall\, x\in(-M,M) $$ And this implies that $\arctan x$ is real analytic.
Does there exist any problems in my poof? And any new ideas would be highly appreciated. Thanks in advance!
Your proof doesn't hold. What is $x_0$? Besides, the series $\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ only converges on $(-1,1)$.
You can prove that $\arctan$ is analytic observing that $\arctan'x=\dfrac1{1+x^2}$ which is analytic, since it is the quotient of two analytic functions. And a primtive of an analytic function is always analytic.