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Does $\int_{-1}^1\frac{1}{x}dx$ Converge or diverge?

Is the following a valid proof of convergence?:

$$= \lim_{a\to0+}(\int_{-1}^{-a}\frac{1}{x}dx + \int_{a}^1\frac{1}{x}dx)$$ $$= \lim_{a\to0+}([\ln|x|]_{-1}^{-a} + [\ln|x|]_{a}^{1})$$ $$= \lim_{a\to0+}(\ln|-a| - \ln|a| + \ln|1|-\ln|-1|)$$ $$= \lim_{a\to0+}(\ln|\frac{-a}{a}|)$$ $$= \ln|-1| = \ln{1} = 0$$

Thanks.

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    $\begingroup$ You have computed the inetgral in the Cauchy principal value sense. But the function is not Riemann (or Lebesgue) integrable in the ordinary sense. $\endgroup$ – Kavi Rama Murthy Mar 9 at 23:52
  • $\begingroup$ @KaviRamaMurthy Right so this means it is convergent in the Cauchy Principal value sense? $\endgroup$ – Moosefeather Mar 9 at 23:53
  • $\begingroup$ pitt.edu/~jwheeler/Principal%20Values.pdf $\endgroup$ – Jean Marie Mar 9 at 23:55
  • $\begingroup$ No, the integral is never convergent. It is used to extrapolate values of integrals that do not converge and give them values. It's the same as saying $1 + 2 + 3 + \cdots = -\frac{1}{12}$: of course, the sum does not converge, but according to similar sums we can give this sum a value by analytic continuation. $\endgroup$ – Toby Mak Mar 9 at 23:56
  • $\begingroup$ According to the definition of CPV here the intergal does converge in the CPV sense: en.wikipedia.org/wiki/Cauchy_principal_value $\endgroup$ – Kavi Rama Murthy Mar 10 at 0:00