Different way to see that $\int_{-c}^c \text{d}{x}/x=0$?

Question

The other day, I stumbled across the following integral: \begin{equation} \int_{-c}^c\text{d}{x}\frac{1}{x} \end{equation} with $c$, a positive real number. Now, it seems to me obvious that this integral should yield zero since it is odd with respect to $x=0$. However, how can one prove this ? It seems that the result depends on how one takes the limit for the $x=0$ singularity. For instance, consider: \begin{equation} \int_{-c}^c\text{d}{x}\frac{1}{x}=\lim_{\tilde\epsilon\rightarrow0^+}\int_{-c}^{-\tilde\epsilon}\frac{\text{d}{x}}{x}+\lim_{\epsilon\rightarrow0^+}\int_{\epsilon}^{c}\frac{\text{d}{x}}{x} \end{equation} The result of the latter expression depends on how one takes the $\epsilon,\tilde\epsilon$ limit... I am a bit confused on how to proceed.