Power series representation for $x\arctan(x)$. How?

Question

I am supposed to start with the P.S representation for $\dfrac{1}{1-x}$ and then figure out the P.S representation for $x\tan^{-1}(x)$. I know the process of making it for just the $\tan^{-1}x$. But wouldn't the fact that there is an extra $x$ at the front change everything dramatically? Or am I not getting something?

Can I just do the same process as for $\tan^{-1}(x)$ and then just add an extra $x$ after sigma?

Thanks!

Answer 1

Recall that $$\arctan x=\int_0^x\frac{dt}{1+t^2}$$ We also know that, for $|t|<1$, $$\frac1t=\sum_{k\geq0}(1-t)^k$$ So we then have that $$\frac1{1+t^2}=\sum_{k\geq0}(-1)^kt^{2k}$$ So for $|x|<1$, $$ \begin{align} \arctan x=&\int_0^x\sum_{k\geq0}(-1)^kt^{2k}dt\\ =&\sum_{k\geq0}(-1)^k\int_0^xt^{2k}dt\\ =&\sum_{k\geq0}(-1)^k\frac{x^{2k+1}}{2k+1}\\ \end{align} $$ Which means that $$x\arctan x=\sum_{k\geq0}(-1)^k\frac{x^{2k+2}}{2k+1}$$

Answer 2

$\tan^{-1}x=\int\frac1{1+x^2}=\int\sum_n(-x^2)^n=\sum_n\int(-x^2)^n=\sum_n(-1)^n\frac{x^{2n+1}}{2n+1}$. Thus $x\cdot\tan^{-1}x=\sum_n(-1)^n\frac{x^{2n+2}}{2n+1}$.