While trying to prove some series (which should be easy to guess) were divergent, I stumbled across the following cute little result. We write \log^{(k)}x = \log \log \log .... \log x with k \logs.

Consider the integral \int_a^b \frac{dt}{t \log t \log \log t ... \log^{(n)}t}.

Then the substitution t=e^s gives this integral as equal to \int_{\log a}^{\log b} \frac{ds}{s \log s \log \log s .... \log^{(n-1)} s}, so making n such substitutions will just reduce this integral to solving \int \frac{dx}{x}, which is well-known.

Advertisements