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I'm wondering how I can prove that this integral is divergent without using the primitive function? $$ \int\limits_0^1 \frac{1}{x}\, dx $$

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  • $\begingroup$ Does "primitive function" mean the antiderivative $\ln$? $\endgroup$ – mvw Feb 4 '15 at 15:48
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    $\begingroup$ By change of variables this is equivalent to the divergence of $\int_1^\infty \frac{1}{y} dy$, which is bounded from below by $\sum_{n=1}^\infty \frac{1}{n+1} =\infty$. $\endgroup$ – Giovanni De Gaetano Feb 4 '15 at 15:52
  • $\begingroup$ @mvw Yes. Even Spivak uses primitive function in this sense. $\endgroup$ – Simon S Feb 4 '15 at 15:53
  • $\begingroup$ Yes, the antiderivate :-) @mvw $\endgroup$ – Louise Feb 4 '15 at 15:53
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For $m\ge 1$, we have $$\int_{1/(m+1)}^{1/m}\frac1x\,\mathrm dx\ge\int_{1/(m+1)}^{1/m}m\,\mathrm dx =\frac1{m+1}$$ hence $$\int_{1/(m+1)}^{1}\frac1x\,\mathrm dx\ge\sum_{k=1}^m\frac1{k+1}$$

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By primitve, I take it that you mean antiderivative. Consider $ 2^n \leq 1/x\ \text{ for all }\ x\in(2^{-(n+1)}, 2^{-n}].$ Therefore $$\int_0^1 \frac{1}{x}\, dx = \sum_{n=0}^\infty \int_{2^{-(n+1)}}^{2^{-n}} \frac{1}{x}\, dx \geq \sum_{n=0}^\infty \int_{2^{-(n+1)}}^{2^{-n}} 2^n\, dx = \sum_{n=0}^\infty \frac{1}{2}$$ which clearly diverges.

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