What's wrong with this wrong derivation of $\ln(-x)$? If we take $\ln(x) = \int_{1}^{x} \frac{1}{y} dy$ as the definition of logarithm, then I am ending with a stupid derivation of $\ln(-x)= - \int_{-x}^{-1} \frac{1}{y} dy$.
From the definition, if I go like this: $$\ln(-x) = \int_{1}^{-x} \frac{1}{y} dy = -(\int_{-x}^{1} \frac{1}{y} dy) = \lim_{\epsilon \rightarrow 0} -(\int_{-x}^{-1} \frac{1}{y} dy + \int_{-1}^{-\epsilon} \frac{1}{y} dy + \int_{-\epsilon}^{\epsilon} \frac{1}{y} dy + \int_{\epsilon}^{1} \frac{1}{y} dy)\\ = \lim_{\epsilon \rightarrow 0} -(\int_{-x}^{-1} \frac{1}{y} dy + \int_{-1}^{-\epsilon} \frac{1}{y} dy + \int_{\epsilon}^{1} \frac{1}{y} dy + \int_{-\epsilon}^{\epsilon} \frac{1}{y} dy )  $$ 
[The middle two terms easily cancel,] $ \lim_{\epsilon \rightarrow 0} -(\int_{-x}^{-1} \frac{1}{y} dy + \int_{-\epsilon}^{\epsilon} \frac{1}{y} dy )$, which is probably equal to $ - \int_{-x}^{-1} \frac{1}{y} dy$.
 A: To use the definition you want, you have to be in a world in which you know how to integrate (and presumably differentiate) but don't know about exponentials. Hardy's Course of Pure Mathematics chapter IX was like that.  Your definition then makes sense for $x>0$ and defines the logarithm on $(0,\infty)$ and lets you prove a ton of stuff, as in Hardy.
You can't readily use your definition for $x\lt0$ because $\int_1^x dy/y=-\int_x^1 dy/y$ is not defined as a standard Lebesgue integral.  That is because, if it existed, it would be the negative of $(\int_x^0+\int_0^1) \dfrac{dy}{y}$, which is the sum of $-\infty$ (the first integral) and $+\infty$ (the second integral).
Also, I get the impression that $\log -1$ is a number $z$ such that $\exp(z)=-1$, for example $z=i(2n+1)\pi$ for some $n\in\Bbb{Z}$, and my imagination fails to see a way to get complex numbers out of your integral.
It is possible I guess that you don't know why $\int_0^1 dy/y=+\infty$.  This is usually proved by saying that it's $\lim_{a\to0+}\int_a^1 dy/y=\lim_{a\to0+} (\log 1 - \log a)$, but if you really and truly don't know anything about logs, then that's not a good proof for you.  Maybe you know that $\sum_{n=1}^\infty 1/n$ diverges, and can base a proof on that?
