Integral of Reciprocal is Divergent

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  • $\displaystyle \int_1^n \frac {\d x} x \to +\infty$ 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 $+\infty$.



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.


Proof of second part

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

Put $x = \dfrac 1 z$.


\(\displaystyle \int_\gamma^1 \frac {\d x} x\) \(=\) \(\displaystyle \int_{1 / \gamma}^1 \frac {-z} {z^2} \rd z\) $\quad$ 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 $\gamma \to 0^+$.


Also see