Why isn't the logarithm defined as its absolute value? As Apostol observes in Calculus, Vol. 1, the function
$$
L(x) = \ln|x| + C
$$
is the unique such function defined on $\mathbf{R} \setminus \{0\}$ respecting:


*

*$L(xy) = L(x) + L(y)$

*$L(1) = 0$

*$\int_{1}^{|x|} \frac{1}{t} \mathrm{d}t$ on $L(x)$'s domain


Since this function appears to be more general than $\ln(x)$ and is differentiable everywhere it's defined following from (3), why isn't $\ln(x)$ defined as $L(x)$? Put differently, why isn't $\ln(x)$ real for negative $x$, since it would apparently simplify the solutions to many integrals of practical interest? Does it have something to do logarithm being defined as an inverse to the exponentiation $\mathrm{e}^{x}$, which is sinusoidal for imaginary $x$ by Euler's formula?
Please excuse my lack of background in complex analysis.
Edit: the original version of this question neglected the constant $C$, and finding its value is the key to answering this question properly.
 A: Since it appears no one wants to supply an official answer to this question, perhaps because it's seen as too elementary, I'll attempt to answer it myself in a very obvious and elementary fashion for the sake of completeness and in hopes that it helps others as this has helped me. Sangchul Lee's comment gave me enough in the way of a hint to finish the proof.
First, assume that $\ln(x)$ can be defined for all $x \in \mathbf{R} \setminus \{0\}$ while satisfying properties 1 through 3 of the question. Now take $x$ to be strictly positive, and observe that $\ln(-x) = \ln((-1) x)$ simply from the field axioms, which by property 1 means:
$$\ln(-x) = \ln((-1) x) = \ln(-1) + \ln(x)$$
Thus to define $\ln(-x)$ for positive $x \in \mathbf{R}^{+}$, it suffices to define $\ln(x)$, which is already accomplished, and to define $\ln(-1)$.
We've several choices for this, but since we've defined $y = \ln(x)$ for positive $x$ as satisfying $\mathrm{e}^{y} = x$, we should see if this generalizes. If it does, then $y = \ln(-1)$ must satisfy $e^{y} = -1$.
By Euler's formula, which can be shown by separating a convergent infinite sum into real and imaginary terms, we have for any $\theta \in \mathbf{R}$:
$$\mathrm{e}^{\mathrm{i} \; \theta} = \cos(\theta) + \mathrm{i} \; \sin(\theta)$$
For a complex $c$ with real and imaginary components $\Re(c) = a \in \mathbf{R}$ and $\Im(c) = b \in\mathbf{R}$, we then have:
$$\mathrm{e}^{c} = \mathrm{e}^{a + \mathrm{i} \; b} = \mathrm{e}^{a} \mathrm{e}^{\mathrm{i} b} = \mathrm{e}^{a}(\cos(b) + \mathrm{i} \; \sin(b)) = \mathrm{e}^{a}\cos{b} + \mathrm{i} \; \mathrm{e}^{a}\sin{b}$$
Here, note carefully that since $a$ is real, $\mathrm{e}^{a}$ is also real; and since $b$ is real, $\cos(b)$ and $\sin(b)$ are real. This means that $\mathrm{e}^{a}\cos{b}$ is real while $\mathrm{i} \; \mathrm{e}^{a}\sin{b}$ is imaginary.
Letting this equal $-1$ to solve for $\ln(-1)$ gives:
$$-1 = \mathrm{e}^{a + \mathrm{i} \; b} = \mathrm{e}^{a} \mathrm{e}^{\mathrm{i} b} = \mathrm{e}^{a}(\cos(b) + \mathrm{i} \; \sin(b)) = \mathrm{e}^{a}\cos{b} + \mathrm{i} \; \mathrm{e}^{a}\sin{b} \quad (\diamond)$$
Now observe that $\mathrm{i}^2 = -1$, so that $i = (-1)^{1/2}$, letting us raise each side of $(\diamond)$ to $1/2$ (using $c/2 = a/2 + \mathrm{i} (b / 2)$) in order to write:
$$\mathrm{i} = \mathrm{e}^{c/2} = \mathrm{e}^{a/2}\cos\left(\frac{b}{2}\right) + \mathrm{i} \; \mathrm{e}^{a/2}\sin\left(\frac{b}{2}\right)$$
As $\mathrm{i}$ is imaginary, each side of the equation has no real part. This implies that:
\begin{equation}
\mathrm{i} =\mathrm{i} \; \mathrm{e}^{a/2}\sin\left(\frac{b}{2}\right) \Rightarrow 1 = \mathrm{e}^{a/2} \sin\left(\frac{b}{2}\right) \quad (\star)
\end{equation}
\begin{equation}
0 = \mathrm{e}^{a/2}\cos\left(\frac{b}{2}\right) \quad (\star \star)
\end{equation}
Since there is no real $\frac{a}{2}$ for which $\mathrm{e}^{a/2} = 0$, we can divide $\mathrm{e}^{a/2}$ out of $(\star \star)$ to write:
\begin{equation}
0 = \cos\left(\frac{b}{2}\right)
\end{equation}
From trigonometry, we know the only two solutions to this equation modulo $2 \pi$ are $b/2 = \pi/2$ and $b/2 = 3\pi/2$.
Since $\mathrm{e}^{a/2} > 0$ while $\sin(\pi / 2) = 1$ but $\sin(3 \pi /2) = -1$, both $(\star)$ and $(\star \star)$ are only satisfied for:
$$e^{a/2} = 1 \Rightarrow a / 2 = 0 \Rightarrow a = 0$$
$$b / 2 = \pi / 2 \Rightarrow b = \pi$$
Substituting these back into $(\diamond)$ gives:
$$-1 = \mathrm{e}^{0 + \mathrm{i} \pi} = \mathrm{e}^{\mathrm{i} \pi}$$
Thus, the only way to define $y = \ln(-x)$ for positive $x \in \mathbf{R}^+$ such that $\mathrm{e}^{y} = -x$ and assumptions 1, 2, and 3 are satisfied is to define $\ln(-1) = \mathrm{i} \pi$, from which it follows that $\ln(-x) = \mathrm{i} \pi + \ln(x)$. Since $\mathrm{i} \pi$ is a constant (namely the only sensible constant $C$ on negative reals in my revised question, and $C = 0$ for positive reals), it vanishes from all definite integrals of $\ln(x)$ on negative reals, giving us property 3. Properties 1 and 2 are assumed in finding this answer.
