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.

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    $\begingroup$ What is your background? $\endgroup$ – Will Jagy Jan 6 '19 at 3:38
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    $\begingroup$ This is a divergent improper integral. WA might give you the principal value, not the exact value. $\endgroup$ – xbh Jan 6 '19 at 3:39
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    $\begingroup$ @EstMayhem I would look into the concept of the Cauchy principal value. This integral is described in the examples. $\endgroup$ – Tyberius Jan 6 '19 at 3:40
  • $\begingroup$ @Tyberius thank you! That's gonna help. $\endgroup$ – Est Mayhem Jan 6 '19 at 3:45

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.$$\begin{align*}\int\limits_{-1}^1\frac {\mathrm dx}x & =\lim\limits_{\varepsilon\to0}\left[\int\limits_{-1}^{-\varepsilon}\frac {\mathrm dx}x+\int\limits_{\varepsilon}^1\frac {\mathrm dx}x\right]\\ & =\lim\limits_{\varepsilon\to0}\Bigr[\log|-\varepsilon|-\log|-1|+\log 1-\log(\varepsilon)\Bigr]\\ & =\lim\limits_{\varepsilon\to0}\log\left(\frac {\varepsilon}{1}\frac 1{\varepsilon}\right)\\ & =\log 1\\ & =0\end{align*}$$

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    $\begingroup$ The first sentence is not correct. The reason the integral is divergent is not because $1/x$ is undefined at $x=0$, it's because a computation shows that it's divergent. $\endgroup$ – Hans Lundmark Jan 6 '19 at 9:33
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    $\begingroup$ Also, you need to take $\epsilon \to 0^+$ and use $\log|x|$ as your antiderivative instead of $\log(x)$. Your computation with $\log(-1)$ is clearly nonsense. $\endgroup$ – Hans Lundmark Jan 6 '19 at 9:35
  • $\begingroup$ @HansLundmark: $\log x$ as antiderivative of $1/x$ is correct for $x$ in the complex plane. The subtraction $\log(-\varepsilon)-\log(-1)$, is real, although each of the two logs is complex. This way of doing it may be inadvisable for beginners, but it is not "clearly nonsense". $\endgroup$ – GEdgar Jan 6 '19 at 13:18
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    $\begingroup$ @GEdgar: I know that. But in the context of trying to explain things about real integrals to someone whose background you don't know (a calculus student perhaps?), I don't think it makes any sense at all. And the log laws ($\log(zw)=\log z + \log w$, etc.) are not valid without restrictions in the complex domain, so the step that follows definitely needs to be carefully justfied. At the very least, one should specify what branch of the complex log that is used. Just writing out a formal computation like that without any explanation doesn't prove anything. $\endgroup$ – Hans Lundmark Jan 6 '19 at 13:35
  • $\begingroup$ @GEdgar: And if one uses the complex logarithm in an answer like this, at least one ought to say so! With no mention of that, it's natural to assume that it's the usual real logarithm, in which case "log(-1)" is indeed nonsense. $\endgroup$ – Hans Lundmark Jan 6 '19 at 13:43

This cannot be proven rigorously because it is not technically true.

The Cauchy principal value of your integral is $0$, but if you go back to the definition of the improper Riemann integral you will find that this quantity is undefined.

  • $\begingroup$ What if this a Lebesgue integral? $\endgroup$ – Est Mayhem Jan 6 '19 at 3:45
  • $\begingroup$ I got it. Thanks for the reply! $\endgroup$ – Est Mayhem Jan 6 '19 at 3:49
  • $\begingroup$ @EstMayhem Well I don't know who replied, but you are welcome? The answer is that the function $1/x$ is not Lebesgue integrable on this interval. $\endgroup$ – ItsJustLogicBro Jan 6 '19 at 3:51
  • $\begingroup$ I meant your initial reply for the question itself :) Thanks. $\endgroup$ – Est Mayhem Jan 6 '19 at 4:36
  • $\begingroup$ The Lebesgue integral also does not exist. $\endgroup$ – GEdgar Jan 6 '19 at 13:19

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