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.

#Calculus

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

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

Hamza A - 4 years, 10 months ago

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

Priyamvad Tripathi - 4 years, 10 months ago

You need to know that log of imaginary number is defined

Dev Rajyaguru - 4 years, 10 months ago

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

Priyamvad Tripathi - 4 years, 10 months ago

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.

Abhi Kumbale - 4 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$