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

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 $\displaystyle \sum_{n=1}^\infty \frac 1 n$ 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 $+\infty$.

$\blacksquare$

Proof 2 of first part

From the definition of natural logarithm (or from Equivalence of Logarithm Definitions):

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \ln x\)

\(=\)

\(\displaystyle \int_1^x \dfrac 1 t \ \mathrm dt\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

The result follows from Logarithm Tends to Infinity.

$\blacksquare$

Proof of second part

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

Put $x = \dfrac 1 z$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \int_\gamma^1 \frac {\mathrm dx} x\)	\(=\)	\(\displaystyle \int_{1 / \gamma}^1 \frac {-z} {z^2} \ \mathrm dz\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integration by Substitution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_1^{1 / \gamma} \frac {\mathrm dz} z\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

From the above result:

$\displaystyle \int_1^{1 / \gamma} \frac {\mathrm dz} z \to +\infty$

as $\gamma \to 0^+$.

$\blacksquare$

Also see

• Logarithm Tends to Infinity

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 13.33$