Value of $\log(x)-\log(-x) $ According to wolframalpha $\log(a)-\log(-a) = i\pi$ for $a<0$ and $-i\pi$ for $a>0$
This is the integral $\int_{-a}^a \frac{1}{x}dx$
I guess we have to be smart with complex integration then...?
Still, i can not fully fanthom how a real integral can have complex output though...
Has anyone enlighting ideas?
 A: Since the integral clearly does not exist, there is no problem. 
Integrals of real functions will continue to give real values.
To see that the integral does not exist even with generalized integral, write:
$$\int_{-a}^a \frac{1}{x} dx =\int_{-a}^{-\epsilon} \frac{1}{x} dx + \int_{\delta}^a \frac{1}{x} dx = \log(\epsilon/a) - \log(\delta/a)$$
Since this can get arbitrary values for any choice of $\epsilon(\delta)$, the generalized integral isn't defined.
A: However the @rlgoronma's answer light the way, but note that the function $f(x)=1/x$ has a discontinuity point at the origin and I don't think that improper integrals can help you.
A: As soon as you are envisaging $\log(-x)$ for some $x>0$ you cannot make sense of it unless you are transcending into the complex domain. There the function $\log$ takes values not in ${\mathbb C}$, but in ${\mathbb C}/(2\pi i{\mathbb Z})$. This means that $\log z$ is defined only "up to an additive multiple of $2\pi i\>$"; it is given by
$$\log z=\log|z| + i\arg (z)\qquad(z\ne0)\ ,$$
where $\arg (z)$ is the "polar angle" of $z$ up to a multiple of $2\pi$. As
$|-z|=|z|$ and $\arg(-z)=\arg (z)+[\pi]$ for all $z\ne0$ we find that
$$\log(z)-\log(-z)= i[\pi]=\{\pi i +2k\pi\>i\ |\ k\in{\mathbb Z}\}\qquad(z\ne0)\ ,$$
and there is no way out of this indeterminacy.
To make things more convenient one introduces the principal value of $\log$, called ${\rm Log}$. This function is defined in ${\mathbb C}$ minus the negative real axis und is equal to the representant of $\log$ having imaginary part in $\ ]{-\pi},\pi[\ $. Unfortunately we cannot make use of this escape when $z=x>0$, because $-z$ then lies on the negative real axis.
