Integral of Reciprocal is Divergent/To Zero

Theorem

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

Thus the improper integral $\ds \int_{\to 0^+}^1 \frac {\d x} x$ does not exist.

Put $x = \dfrac 1 z$.

Then:

$\ds \int_\gamma^1 \frac {\d x} x$

$=$

$\ds \int_{1 / \gamma}^1 \frac {-z} {z^2} \rd z$

$\ds $

$=$

$\ds \int_1^{1 / \gamma} \frac {\d z} z$

But as $\gamma \to 0+$, we have that $\dfrac 1 \gamma \to +\infty$.

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

as $\dfrac 1 \gamma \to +\infty$.

The result follows.

$\blacksquare$