the values of ln for negative x are not the values of the anti-derivative of 1/x for negative x

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Posted on 2011-05-10

So I know derivative of ln(x) = 1/x, but the anti-derivative of 1/x = ln(|x|), why the absolute value? It is possible to have negative values for x in ln, so why isn't it just x?

5 Comments

means that for x<0

F(x) = ln(-x)

F'(x) = (-1) / (-x) = 1/x

so the anti-derivative is shown to be valid

but also of course

d/dx(ln(x)) = 1/x

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in the real numbers ln(x) is not defined for x<0, but in complex numbers

-x = x e^(pi i)

since e^(pi i) = -1

so for example

ln(-5) = ln(5 e^(pi i)) = ln(5) + pi i

and in general

ln(-x) = ln(x) + pi i

therefore the difference between ln(-x) and ln(x) is a constant which can be absorbed into a constant of integration, and ultimately would not affect a definite integral

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integral ln(x) is found by considering ln(x) = 1 ln(x) , then using integration by parts to give

x ln(x) - x

Im sure you will find the following link interesting.

http://en.wikipedia.org/wi

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