Negative number in logarithm

Question

I was given a problem to evaluate the following integral, $$\int\limits_{-4}^{-1}\dfrac{1}{x}dx$$

I know the fact that integration of $1/x$ is $\log(x)$ but how can we compute negative logarithm? I mean to say that since definite integral refers to the area under the curve so how can it be even defined?

I tried the following method: $$\int\limits_{-4}^{-1}\dfrac{1}{x}dx = \log(-1) - \log(-4)$$ $$\implies \log\left(\dfrac{-1}{-4}\right) = -\log(4)$$

Is it correct? I don't think that the answer should be $-\log(4)$ because there is no graph for the negative logarithm. Is there any need of complex logarithm?

Nope. The indefinite integral (antiderivative) of $\frac{1}{x}$ (on any interval not containing $0$) is $\log(|x|)+C$, not $\log(x)+C$. — , Jan 02 '22 at 13:54
@StinkingBishop Oh, I didn't know that, I'm just a newcomer to integrals. Thanks for the help! :)) — , Jan 02 '22 at 13:56
Check it out. If $f(x)=\log(-x)$ on $x<0$, then $\frac{df}{dx}=\frac{1}{-x}\cdot (-1)$ (chain rule) $=\frac{1}{x}$. — , Jan 02 '22 at 13:57
In fact, the $C$ in @StinkingBishop's comment can take different values either side of $x=0$, so your definite integral is only well-defined because $0$ isn't between its limits. We say $C$ is locally constant. — J.G., Jan 02 '22 at 14:09
$\int\limits_{-4}^{-1}\dfrac{1}{x}dx =- \int\limits_{1}^{4}\dfrac{1}{y}dy$ — Henry, Jan 02 '22 at 14:20

score 1 · Accepted Answer · answered Jan 02 '22 at 14:08

Negative logarithms do exist, just not in the real numbers. The complex log function yields the same value as the log function, but on negative numbers it adds a complex component, but which is constant!

As a commenter noted, most books put the integral of $\frac{1}{x}$ as $\ln(|x|)$ to prevent going outside the domain. However, if you allow for complex functions, $\ln(x)$ actually still works. In fact, they are equivalent, if you allow $C$ to be complex, and allow for the fact that you probably shouldn't integrate across the non-smooth point at $x = 0$ (thus you may have different constants of integration).

For instance, for the natural logarithm, logarithms of negative numbers get $i\,\pi$ added to their positive equivalent. Therefore, if, between 0 (the non-smooth point) and $-\infty$, the constant of integration included $-i\,\pi$, then the two definitions would give you the same result.

What is a “complex component”? – Radial Arm Saw Jan 02 '22 at 15:49 — Radial Arm Saw, Jan 02 '22 at 15:49

Negative number in logarithm

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