Integral of Reciprocal is Divergent

Theorem

 * $$\int_1^n \frac {dx} x \to +\infty$$ as $$n \to + \infty$$;
 * $$\int_\gamma^1 \frac {dx} x \to -\infty$$ as $$\gamma \to 0^+$$.

Thus the improper integrals $$\int_1^{\to +\infty} \frac {dx} x$$ and $$\int_{\to 0^+}^1 \frac {dx} x$$ do not exist.

Proof

 * $$\int_1^n \frac {dx} x \to +\infty$$ as $$n \to + \infty$$:

From Sum of Reciprocals is Divergent, we have that $$\sum_{n=1}^\infty \frac 1 n$$ diverges to $$+\infty$$.

Thus from the Euler-Maclaurin Summation Formula (also known as the Integral Test), $$\int_1^n \frac {dx} x \to +\infty$$ also diverges to $$+\infty$$.


 * $$\int_\gamma^1 \frac {dx} x \to -\infty$$ as $$\gamma \to 0^+$$:

Put $$x = \frac 1 z$$.

Then $$\int_\gamma^1 \frac {dx} x = \int_{1 / \gamma}^1 \frac {-z dz} {z^2}$$

$$ $$

From the above result, $$\int_1^{1 / \gamma} \frac {dz} {z} \to \infty$$ as $$\gamma \to 0^+$$.