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I've always been taught that when integrating a function of the form $f'(x)/f(x)$ to put an absolute value around the argument of the resulting logarithm. For example:

$$\int\frac1{x}\mathrm dx = \log{|x|} + c$$

The reason provided was that 'logarithms aren't defined for negative numbers', it seems a bit like cheating to me to just throw absolute values around the argument. Furthermore, I thought of a case where this would actually produce the wrong result;

$$\int_{-1}^1\frac1{x}\mathrm dx = \log|1| - \log|-1| = 0$$

However, the correct way should be this:

$$\int_{-1}^1\frac1{x}\mathrm dx = \log(1) - \log(-1) = 0 - i\pi = -i\pi$$

Edit: I may be wrong, but the integral above, ignoring the singularity (sorry couldn't think of a better example to illustrate my point with -1 and changing it now would make people's answers and comments seem off-topic), should be correct due to Euler's identity:

$e^{i\pi} = -1 \implies \log(-1) = i\pi$

Could someone please provide a better explanation?

Thanks

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    For $\int_{-1}^1\frac1{x}\mathrm dx$, you do remember that your integrand's singular in the center of the integration interval? That being said, there's a way to interpret the integral so that the result 0 makes sense, due to Cauchy...2011-07-17
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    One way to think about it: the reciprocal function is odd, so $\frac1{-x}=-\frac1{x}$. Ignoring the singularity, the "area" on the left of the vertical axis is precisely the negative of the area on the right of the axis, so if you add them together...2011-07-17
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    You should go back to the geometric concept of the integral. If you consider the left section of the reciprocal function, wouldn't you consider it off-base that the area bounded by the function and the axis gives a complex result?2011-07-17

3 Answers 3