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So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$

Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:

$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies \begin{cases} A+B=0 \\[2ex] A-B=\frac{1}{i}=-i \end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$

So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$

After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$

$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$

now my question is how can i prove that without using the integral

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    $\begingroup$ Also see math.stackexchange.com/questions/1823688/… $\endgroup$ – Nosrati Sep 2 '17 at 17:31
  • $\begingroup$ @MyGlasses thx for pointing who 2 out, i get how they done it in those 2, the only problem is the $C=-i\pi$ i got, why it doesn't appear at the first one you sent? $\endgroup$ – ℋolo Sep 2 '17 at 17:52
  • $\begingroup$ I'm sorry. I was confused those and worked on second. I think the title of that post is wrong and your proof shows it! $\endgroup$ – Nosrati Sep 2 '17 at 18:29
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By definition of $\ln$ for complex numbers: $$\ln(x+i)=\ln|x+i|+i\operatorname{Arg}(x+i)$$ where $\operatorname{Arg}$ is the principal value of the $\arg$

It is clear that: $$\operatorname{Arg}(x+i)=\begin{cases}\arctan\frac{1}{x}&\text{ if }x\ge0\\\pi+\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$

Analogously:

$$\ln(x-i)=\ln|x-i|+i\operatorname{Arg}(x-i)$$ In this case it is:

$$\operatorname{Arg}(x-i)=\begin{cases}-\arctan\frac{1}{x}&\text{ if }x\ge0\\-\pi-\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$

Moreover observe that: $|x+i|=|x-i|$

Putting everything together you get:

$\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)=\begin{cases}\stackrel{x\ge0}=\frac{i}{2}(2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}+\frac{\pi}{2}=\\\stackrel{x<0}=\frac{i}{2}(2i\pi+2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}-\frac{\pi}{2}=\end{cases}=\arctan x$

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  • $\begingroup$ You made a mistake at the last part, you wrote $\ln(x+i)-\ln(x+i)$ and not $\ln(x+i)-\ln(x-i)$ $\endgroup$ – ℋolo Sep 2 '17 at 19:24
  • $\begingroup$ @Holo Thanks. I'm going to edit the answer. $\endgroup$ – trying Sep 2 '17 at 19:32
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Perhaps using Euler? $$ z= \tan w = \frac{\sin w}{\cos w} = \frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{iw}} = \frac{1}{i} \frac{e^{2iw}-1}{e^{2iw}+1}$$ so for any $k\in {\Bbb Z}$ $$e^{2iw} = e^{2i(w -k)} = \frac{1+iz}{1-iz}$$ whence $$ w = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right) + k \pi$$ where the choice of $k\in {\Bbb Z}$ depends upon the cut you want. $k=0$ corresponds to standard choices of $\log(1)=0$ and $\arctan 0=0$.

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You found $$\arctan x=\dfrac{i}{2}\Big(\ln(x+i)-\ln(x-i)-i\pi\Big)=\dfrac{i}{2}\ln\dfrac{x+i}{x-i}+\dfrac{\pi}{2}=-\dfrac{i}{2}\ln\dfrac{x-i}{x+i}+\dfrac{\pi}{2}$$ let $x=\dfrac1z$ so $$\arctan\dfrac1z=-\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}+\dfrac{\pi}{2}$$ and then $$\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}=\dfrac{\pi}{2}-\arctan\dfrac1z=\arctan z$$ as you want!

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Here is another way to look at this: consider that $z=|z|e^{i\theta}$ and that $z^*=|z^*|e^{-i\theta}$. Then

$$ \frac{|z|}{|z^*|}=e^{2i\theta}\\ \theta=\frac{1}{2i}(\ln |z|-\ln |z^*|) $$

Now, let $z=x+i$, then

$$ \theta=\cot^{-1}x=\frac{1}{2i}(\ln |x+i|-\ln |x-i|) $$

Now,

$$ \cot^{-1}x+\tan^{-1}x=\frac{\pi}{2} $$

so that

$$ \begin{align} \tan^{-1}x &=-\frac{1}{2i}(\ln |x+i|-\ln |x-i|)+\frac{\pi}{2}\\ &=\frac{i}{2}(\ln |x+i|-\ln |x-i|-i\pi) \end{align} $$

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