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on the one hand, clearly $$\int \frac{1}{x^2+1}dx=\arctan(x)+c$$ on the other hand using partial fraction decomposition $$\frac{1}{x^2+1}=\frac{A}{x+i} +\frac{B}{x-i}$$ with $A=1/(-2i)$ and $B=1/(2i)$ leading to $$\int \frac{1}{x^2+1}dx=\frac{-1}{2i}\ln(x+i)+\frac{1}{2i}\ln(x-i)$$ leading to the conclusion that [up to a constant..] $$\arctan(x) = \frac{-1}{2i}\ln(x+i)+\frac{1}{2i}\ln(x-i)$$ leading to two thoughts.. 1) is there a flaw in above reasoning? 2) is there another way to verify that $$\arctan(x) = \frac{-1}{2i}\ln(x+i)+\frac{1}{2i}\ln(x-i) + c$$ other than taking derivatives on both sides..

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    $\begingroup$ You might be interested in this $\endgroup$
    – L. F.
    Aug 29, 2013 at 18:20
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    $\begingroup$ you have to drop the absolute values from the logarithms $\endgroup$
    – user8268
    Aug 29, 2013 at 18:21
  • $\begingroup$ thank you L.F. exactly what i was looking for. Very nice! $\endgroup$
    – userX
    Aug 29, 2013 at 18:22
  • $\begingroup$ For alternate proofs of the relationships in L.F.'s link, see this related question. $\endgroup$
    – Meow
    Sep 2, 2013 at 9:28

4 Answers 4

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The form you got (up to an absolute value) is indeed correct. It may look confusing (what are imaginary numbers doing here?) but it is correct. To see this geometrically, construct a right triangle in the complex plane with base (real) $x$ and height of $i$.

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    $\begingroup$ That "up to absolute value" is important. Integral of $1/(x+i)$ is $\log(x+i)+C$ and NOT $\log|x+i|$. NOT NOT NOT. $\endgroup$
    – GEdgar
    Aug 29, 2013 at 18:42
  • $\begingroup$ i shall edit it now.. thx GEdgar $\endgroup$
    – userX
    Aug 29, 2013 at 18:46
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Hint: let $$x=\tan u$$ so $$\arctan(\tan u)=u:0<u<\pi/2$$

then put $$\tan u=\frac{e^{ui}+e^{-ui}}{i(e^{ui}-e^{-ui})}$$

so you get $$u=\frac{1}{2i}\ln(\frac{(\frac{e^{ui}+e^{-ui}}{i(e^{ui}-e^{-ui})})-i}{(\frac{e^{ui}+e^{-ui}}{i(e^{ui}-e^{-ui})})+i})+c$$

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${\large x > 0}$: $$ {\rm F}\left(x\right) = \int_{0}^{x}{{\rm d}z \over z^{2} + 1} = {1 \over 4{\rm i}}\int_{-x}^{x}{{\rm d}z \over z - {\rm i}} - {1 \over 4{\rm i}}\int_{-x}^{x}{{\rm d}z \over z + {\rm i}} = {1 \over 4{\rm i}}\int_{-x -{\rm i}}^{x - {\rm i}}{{\rm d}z \over z} - {1 \over 4{\rm i}}\int_{-x + {\rm i}}^{x + {\rm i}}{{\rm d}z \over z} $$

Define $z^{-1} = \left\vert z\right\vert^{-1}{\rm e}^{-{\rm i}\phi\left(z\right)}$ where $z \in \left\lbrace z' \in {\mathbb C}\ \ni\ z' \not=x\,,\ x\geq 0\right\rbrace$ and $0\ <\ \phi\left(z\right)\ <\ 2\pi$.

\begin{align} \int_{-x - {\rm i}}^{x - {\rm i}}{{\rm d}z \over z} &= -\int_{-1}^{0^{+}}{{\rm i}\,{\rm d}y \over x + {\rm i}y} - \int_{x}^{-x}{{\rm d}x' \over x' - {\rm i}0^{+}} - \int_{0^{+}}^{-1}{{\rm i}\,{\rm d}y \over {\rm i}y} \\[3mm] \int_{-x + {\rm i}}^{x + {\rm i}}{{\rm d}z \over z} &= -\int_{1}^{0^{+}}{{\rm i}\,{\rm d}y \over x + {\rm i}y} - \int_{x}^{-x}{{\rm d}x' \over x' + {\rm i}0^{+}} - \int_{0^{+}}^{1}{{\rm i}\,{\rm d}y \over {\rm i}y} \\[5mm]& \end{align}

\begin{align} & \int_{-x - {\rm i}}^{x - {\rm i}}{{\rm d}z \over z} - \int_{-x + {\rm i}}^{x + {\rm i}}{{\rm d}z \over z} = -{\rm i}\,{\cal P}\int_{-1}^{0^{+}}{{\rm d}y \over x + {\rm i}y} - {\rm i}\,{\cal P}\int^{1}_{0^{+}}{{\rm d}y \over x + {\rm i}y} \\[3mm]& + \int_{-x}^{x}\left({1 \over x' - {\rm i}0^{+}} - {1 \over x' + {\rm i}0^{+}}\right)\,{\rm d}x' + {\cal P}\int_{-1}^{1}{{\rm d}y \over y} \\[3mm]&= -{\rm i}\int_{-1}^{0^{+}} {{\rm d}y \over \sqrt{x^{2} + y^{2}\,}\ {\rm e}^{-{\rm i}\left\lbrack 2\pi - \Theta\left(x,y\right)\right\rbrack}} - {\rm i}\int^{1}_{0^{+}} {{\rm d}y \over \sqrt{x^{2} + y^{2}\,}\ {\rm e}^{-{\rm i}\Theta\left(x,y\right)}} + 2\pi{\rm i} \end{align}

where $\theta\left(x,y\right) \equiv \arctan\left(\left\vert y\right\vert/x\right)$

\begin{align} & \int_{-x - {\rm i}}^{x - {\rm i}}{{\rm d}z \over z} - \int_{-x + {\rm i}}^{x + {\rm i}}{{\rm d}z \over z} = -4{\rm i}\int^{1}_{0^{+}} {\cos\left(\theta\left(x,y\right)\right)\over \sqrt{x^{2} + y^{2}\,}}\,{\rm d}y + 2\pi{\rm i} = -4{\rm i}\int^{1}_{0^{+}}{x \over y^{2} + x^{2}\,}\,{\rm d}y + 2\pi{\rm i} \\[3mm]&= -4{\rm i}\int_{0}^{1/x}{{\rm d}y \over y^{2} + 1} + 2\pi{\rm i} \\[3mm]&{\large \Longrightarrow}\quad {\rm F}\left(x\right) = -{\rm F}\left(1 \over x\right) + {\pi \over 2} \end{align}

Then, $$ {\rm F}\left(x\right) = {\large\int_{0}^{x}{{\rm d}x \over x^{2} + 1} = \arctan\left(x\right)\,, \qquad x > 0} $$

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You're trying to show ${\displaystyle \arctan(x) = -{1 \over 2i}\log{x + i \over x - i}}$ plus a constant. Notice that ${\displaystyle {x + i \over x - i}}$ is a complex number of modulus $1$, so is of the form $e^{i\theta}$ for some $\theta$. So you want to show that $\arctan(x) = -{1 \over 2}\theta$ plus some constant. Now draw $x + i$ and $x - i$ on the complex plane and create two triangles, the first with vertices $(0,0), (x,0),$ and $(x,1)$ and the second its reflection across the $x$ axis, with vertices $(0,0), (x,0),$ and $(x,-1)$.

Then $\theta$ is the angle from $(x,1)$ to $(0,0)$ to $(x,-1)$, and from your picture you should be able to see that $\cot({\theta \over 2}) = x$. So $${\theta \over 2} = {\rm arccot}\,x$$ $$= {\pi \over 2} - \arctan x$$ So what you get is actually $$\arctan x = {\pi \over 2} - {\theta \over 2}$$

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