Integral of Reciprocal is Divergent

Theorem

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

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

Proof 1 of first part

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

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

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

Proof 2 of first part
From the definition of natural logarithm (or from Equivalence of Logarithm Definitions):

The result follows from Logarithm Tends to Infinity.

Proof of second part

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

Put $x = \dfrac 1 z$.

Then:

From the above result:
 * $\displaystyle \int_1^{1 / \gamma} \frac {\mathrm dz} z \to +\infty$

as $\gamma \to 0^+$.

Also see

 * Logarithm Tends to Infinity