Test the improper integral ∫ 1 / x (log x)s for convergence

Test the following improper integral for convergence:

$\int_2^{\infty} \frac{dx}{x (\log x)^s}.$

The integral converges if $s < 1$ and diverges for $s \geq 1$ .

Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution $u = \log x$ ) for $s \neq 1$ :

$\begin{align*} \int \frac{1}{x (\log x)^s} \, dx &= \int \frac{1}{u^s} \, du \\ &= \frac{u^{1-s}}{1-s} \\ &= \frac{(\log x)^{1-s}}{1-s}. \end{align*}$

But then the limit

$\lim_{x \to +\infty} \frac{(\log x)^{1-s}}{1-s}$

is finite for $s > 1$ and diverges for $s \leq 1$ (since $\log x \to \infty$ as $x \to \infty$ and so the limit diverges when $1-s > 0$ and converges for $1-s < 0$ ). Hence, the integral

$\int_2^{\infty} \frac{dx}{x (\log x)^s}$

converges for $s > 1$ and diverges for $s < 1$ . In the case that $s = 1$ the indefinite integral is

$\int \frac{1}{x \log x} \, dx = \log (\log x)$

and $\log (\log x) \to +\infty$ as $x \to +\infty$ , so the integral divers for $s = 1$ as well. Therefore, we have the integral converges for $s > 1$ and diverges for $s \leq 1. \qquad \blacksquare$

Stumbling Robot

Test the improper integral ∫ 1 / x (log x)^s for convergence

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