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

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

Proof 1
From Sum of Reciprocals is Divergent, we have that $\displaystyle \sum_{n=1}^\infty \frac 1 n$ 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
For ease of presentation, we use:


 * $\ln x = \displaystyle \int_1^x \frac 1 t \ \mathrm dt$

Lemma

 * $\ln 2 > \dfrac 1 2$

Proof of lemma
We have that the natural logarithm is continuous.

Hence the Mean Value Theorem for Integrals can be applied.

There exists some $k \in \left[1..2\right]$ such that:


 * $\displaystyle \ln 2 = \frac 1 k \left({2 - 1}\right) = \frac 1 k$.

This means that:


 * $1 \le k \le 2$


 * $\implies 1^{-1} \ge k^{-1} \ge 2^{-1}$


 * $\implies 1 \ge \ln 2 \ge \dfrac 1 2$

Main result
From the definition of infinite limits at infinity, our assertion is:


 * $\forall M > 0 \ \exists N > 0 : x > N \implies \ln x > M$.

As $x \to +\infty$, we will restrict out attention to sufficiently large $M$.

Now, because the natural logarithm is increasing, for sufficiently large $M$:


 * $x > 2^{2M} \implies \ln x > \ln 2^{2N}$

From the Laws of Logarithms:


 * $\quad \ge 2M \cdot \dfrac 1 2 = M$

So:


 * $\forall M \ge a > 0 \ \exists N > 0: x > N = 2^{2M} \implies \ln x > M $

where $a$ is some large number.

Hence the result, by the definition of infinite limits at 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 $\gamma \to 0^+$.