# Integral of Reciprocal is Divergent

## Contents

## 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

### 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 $+\infty$.

$\blacksquare$

#### Note

The integral test works in both directions.

That is, it can also be used to show that Harmonic Series is Divergent *based* on Integral of Reciprocal is Divergent.

This could lead to a circular proof.

Hence, if the integral test is used here, it should **not** be used to prove Harmonic Series is Divergent.

Instead, use one of the other proofs.

### Proof 2 of first part

From the definition of natural logarithm:

\(\displaystyle \ln x\) | \(=\) | \(\displaystyle \int_1^x \dfrac 1 t \rd t\) | $\quad$ | $\quad$ |

The result follows from Logarithm Tends to Infinity.

$\blacksquare$

### Proof of second part

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

Put $x = \dfrac 1 z$.

Then:

\(\displaystyle \int_\gamma^1 \frac {\d x} x\) | \(=\) | \(\displaystyle \int_{1 / \gamma}^1 \frac {-z} {z^2} \rd z\) | $\quad$ Integration by Substitution | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int_1^{1 / \gamma} \frac {\d z} z\) | $\quad$ | $\quad$ |

From the above result:

- $\displaystyle \int_1^{1 / \gamma} \frac {\d z} z \to +\infty$

as $\gamma \to 0^+$.

$\blacksquare$

## Also see

## Sources

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