Prove $\int_0^1\frac{1}{x}dx=\infty$

Question

How do I prove the following? $$\int_0^1\frac{1}{x}dx=\infty $$

Answer 1

Alternative approach: assume that $\int_{0}^{1}\frac{dx}{x}=C\in\mathbb{R}^+$. By enforcing the substitution $x\mapsto 2z$ we have

$$ C=\int_{0}^{1}\frac{dx}{x} = \int_{0}^{\frac{1}{2}}\frac{dz}{z} $$ from which it follows that $$ 0 = \int_{\frac{1}{2}}^{1}\frac{dy}{y} \geq \int_{\frac{1}{2}}^{1}\frac{dy}{1} = \frac{1}{2}, $$ contradiction.

Answer 2

$$\int_0^1\frac{1}{x}dx=\left[\ln|x|\right]_0^1=\lim_{x\to 0}\ln(x)=\infty$$

Answer 3

$\int_0^1\frac1x\operatorname{dx}=\lim_{\epsilon\to0^+}\int_{\epsilon}^1\frac1x\operatorname{dx}=\lim_{\epsilon \to0^+}[\ln x]_{\epsilon}^1=\ln1-\lim_{\epsilon\to0^+} \ln\epsilon=0-(-\infty)=\infty $.