# Integration of complex numbers is really complex

$\large \int \dfrac{1}{x^2+1} \, dx = arctanx + k$ but the same integral can be written as
$\large \int \dfrac{1}{x^2-i^2} \, dx$ where$i=\sqrt{-1}$ Using partial fractions and then integrating $\large \int \dfrac{1}{x^2-i^2} \, dx = 1/2i* \ln \frac { x-i }{ x+i } + C$ Prove that their difference is a constant.

1 year, 9 months ago

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Take

x - i =re^(iy)

So taking conjugates both sides,

x + i = re^(-iy)

So,

ln (x-i) - ln (x+i)= ln (re^(iy)) - ln (re^(-iy))= 2iy

Here y= arctan (-1/x) = arctan (x) - pi/2 .

So the integrals differ by a constant. Hope this helps sorry for typing errors.

- 1 year, 8 months ago

You need to know that log of imaginary number is defined

- 1 year, 8 months ago

Yes I came to know it after I posted this note and even proved the same.

- 1 year, 8 months ago

Do you mean $\frac { x-i }{ x+i }$ or $x-\frac { i }{ x+i }$ ? and $\frac{1}{2}i$ or $\frac{1}{2i}$?

- 1 year, 9 months ago

Sir I have corrected it. It was the first one $\frac { x-i }{ x+i }$

- 1 year, 9 months ago