TITLE: How can I prove that $\int _{-1}^{1} \frac{1}{x} dx =0 $
QUESTION [8 upvotes]: According to WolframAlpha $\int _{-1}^{1} \frac{1}{x} dx =0 $. But how can this be proved rigorously? I know that the function is odd, but it's unbounded on $[-1;1]$. Leibniz–Newton formula cannot be applied here either. I would really appreciate some help with this matter.

REPLY [13 votes]: The integral is zero when we consider the principal value of the integral. I assume that this is what Wolfram Alpha confirms.
To prove this, split the integral up over its positive and negative parts and replace the zero with a variable and have it tend towards zero.$$\begin{align*}\int\limits_{-1}^1\frac {\mathrm dx}x & =\lim\limits_{\varepsilon\to0}\left[\int\limits_{-1}^{-\varepsilon}\frac {\mathrm dx}x+\int\limits_{\varepsilon}^1\frac {\mathrm dx}x\right]\\ & =\lim\limits_{\varepsilon\to0}\Bigr[\log|-\varepsilon|-\log|-1|+\log 1-\log(\varepsilon)\Bigr]\\ & =\lim\limits_{\varepsilon\to0}\log\left(\frac {\varepsilon}{1}\frac 1{\varepsilon}\right)\\ & =\log 1\\ & =0\end{align*}$$