Integral of Reciprocal is Divergent

Theorem

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

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

Proof 1 of first part

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

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

Thus from the Integral Test, $\displaystyle \int_1^n \frac {\d x} x \to +\infty$ also diverges to $+\infty$.

Note: while this is a valid proof, it can lead to a cycle, as the integral test works in both directions. Hence, one should either use one of the other proofs for divergence of the Harmonic series (like this one), or use the next proof below for the first part of this theorem.

Proof 2 of first part
From the definition of natural logarithm:

The result follows from Logarithm Tends to Infinity.

Proof of second part

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

Put $x = \dfrac 1 z$.

Then:

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

as $\gamma \to 0^+$.

Also see

 * Logarithm Tends to Infinity