# Integration proof without using primitive function

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

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

## 2 Answers

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 $$\int_{1/(m+1)}^{1}\frac1x\,\mathrm dx\ge\sum_{k=1}^m\frac1{k+1}$$

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.