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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

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