Integration with complex constants

Question

So I saw a youtube video of the following integral, which can be solved by letting u=tanx, then you get

$\int \frac{1}{1+x^2} dx $ = ${tan}^{-1}x + C$

But it should also be solvable through partial fraction decomposition.

$\int \frac{1}{1+x^2}dx$ = $\int {\frac{A}{x-i} + \frac{B}{x+i}dx}$

$\frac{1}{1+x^2}$ = $\frac{A}{x-i}$ + $\frac{B}{x+i}$

$1$ = $A(x+i)$ + $B(x-i)$

$x=i$: $1$ = $2iA$

$A$=$\frac 1{2i}$ = -$\frac {i}{2}$

$x=-i$: $1 = -2iB$

$B$ = $\frac{i}2$

$\int \frac{1}{1+x^2}dx$ = $\int -\frac{i}{2(x-i)}$ + $ \frac{i}{2(x+i)}dx$ = $-\frac{i}2$ln$\lvert {x-i}\rvert$ + $\frac{i}2$$ln\lvert {x+i}\rvert$ + C = $\frac{i}2ln\lvert {x+i}\rvert - ln\lvert {x-i}\rvert$ + C

This should be equal to ${tan}^{-1}x + C$ right?

${tan}^{-1}({1})$ = $\frac\pi4$

$\frac{i}2(ln\lvert {1+i}\rvert - ln\lvert {1-i}\rvert)$ = $\frac{i}2(ln{\sqrt2}-ln{\sqrt2})$=O

So these are obviously not equal, then I thought why must i have absolute value sign in the logarithmic expressions, that rule I know only applies for integarting with real constants.

Thus,

$\int \frac{1}{1+x^2} dx = $$\frac{i}2(ln({x+i}) - ln({x-i}))$ + C = $\frac{i}2ln(\frac{({x+i})^2}{x^2+1}) + C$

Evaluated at x=1 gives us

$\frac{i}2ln(\frac{({1+i})^2}{1^2+1})$ = $\frac{i}2ln(i)$ = $\frac{i}2\cdot \frac{i\pi}{2}$ = $-\frac\pi4$

So, im quite close, but I must have missed something somewhere, and I cant find the mistake. But this triggers two other questions: How would I show that

$\frac{i}2ln(\frac{({x+i})^2}{x^2+1})$ = $\tan^{-1}x$ (or some other logarithmic expression if that expression is where the error lies)

and, why dont I need to take absolute value when integrating

$\int \frac{1}{x+a}dx$

with complex constants? Is the explanation as easy that I can evaluate all logarithmic expressions using complex numbers?

Answer 1

$$e^{xi} = \cos{x}+i\sin{x}$$ $$e^{-xi} = \cos{x}-i\sin{x}$$ $$\sin{x} = \frac{ e^{xi}-e^{-xi} }{2i}$$ $$ \cos{x} = \frac{ e^{xi}+e^{-xi} }{2}$$ $$\tan{x} = \frac{\sin{x}}{\cos{x}} =\frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$ $$\tan{x} = \frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} }$$ Say $ \frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$ $$ \tan{x} = y , x = \arctan{y}$$ $$ \frac{1}{i} \frac{ e^{xi}-e^{-xi} }{e^{xi}+e^{-xi} } = y$$ Say $x = i \log{z}$ $$ \frac{1}{i} \frac{ z^{-1}-z }{z^{-1}+z} = y$$ $$ z^{-1}-z = i y (z^{-1}+z) $$ $$1-z^2 = i y (1+z^2)$$ $$i y z^2+z^2 = 1-i y$$ $$ z = \sqrt{ \frac{ 1-i y}{1+i y} }$$ $$\arctan{y} = i \log{ \sqrt{ \frac{ 1-i y}{1+i y} }}$$ $$\arctan{y} = \frac{i}{2} \log{ \frac{ 1-i y}{1+i y }}$$ $$\arctan{y} = \frac{i}{2} \log{ \frac{ (1-i y)^2}{(1+i y )(1-i y)}}$$ $$\arctan{y} = \frac{i}{2} \log{ \frac{ 1-2y i -y^2}{1+y^2 }}$$