Show that $\frac{d^n}{dx^n}\left[ \frac{1}{1+x^2} \right] = \frac{(-1)^nn!}{(x^2+1)^{\frac{n+1}{2}}}\sin[(n+1)\arctan x]$

Question

We have, $$\frac{1}{1+x^2}=\frac{A}{x-i}+\frac{B}{x+i} $$ From this we get $A=\frac{1}{2i}, B=-\frac{1}{2i}$. So, $$\frac{1}{1+x^2}=\frac{1}{2i}\left[\frac{1}{x-i} + \frac{1}{x+i} \right] $$ Now, \begin{align*}\left(\frac{1}{1+x^2}\right)^{(n)}&=\frac{1}{2i}\left[\left(\frac{1}{x-i} \right)^{(n)} + \left( \frac{1}{x+i}\right)^{(n)} \right] \\ &= \frac{1}{2i}\left[ \frac{(-1)^{n}n!}{(x-i)^{n+1}} - \frac{(-1)^{n}n!}{(x+i)^{n+1}}\right] \\ &= \frac{(-1)^{n}n!}{2i} \left( \frac{(x+i)^{n+1}-(x-i)^{n+1}}{(x^2+1)^{n+1}}\right) \end{align*} Then we have, $$ (x+i)^{n+1}=(x^2+1)^{\frac{n+1}{2}}\left(\cos((n+1)\cdot \mathrm{arccot}x) + i \sin((n+1)\mathrm{arccot}x) ) \right)$$ $$ (x-i)^{n+1}=(x^2+1)^{\frac{n+1}{2}}\left(\cos((n+1)\cdot \mathrm{arccot}(-x)) + i \sin((n+1)\mathrm{arccot}(-x)) ) \right)$$ But this is where I'm stuck, because we are not winning $\arctan$, maybe it is some elementary trygonometry identity that I'm missing.

Answer 1

HINT:

Let $x=r\cos y,1=r\sin y\implies\cot y=x\implies y=\text{arccot}(x)$

See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

$$(x-i)^m=r^m(\cos y-i\sin y)^m=(1+x^2)^{m/2}\left\{\cos(-y)+i\sin(-y)\right\}^m$$

Using De Moivre' Theorem,

$$\dfrac1{(x+i)^m}=\dfrac{(x-i)^m}{(x^2+1)^m}=\dfrac{(1+x^2)^{m/2}(\cos my-i\sin my)}{(x^2+1)^m}$$

Can you take it from here?

Answer 2

Because of $x+i=\sqrt{x^2+1}e^{i \arctan(x)}$ we get $\frac{1}{x+i}=\frac{1}{\sqrt{x^2+1}}e^{-i \arctan(x)}$. Then continue as you have done.