Does this integral converge?
$$ \int_2^\infty \frac{y}{5\ln^2(y)} \frac{d(\frac{\ln(y-1)}{\ln(y)})}{dy} dy$$
Basic Mathematica calculations (up to $\int_2^{10000000}$) suggest it converges to approximately 0.579. But is there any mathematical way to prove it?
Yes. I was wrong before when I said no. (Thanks, Claude Leibovici.)
$$\frac{d(\frac{\ln(y-1)}{\ln(y)})}{dy} = \frac{\frac{\ln y}{y-1} - \frac{\ln(y-1)}{y}}{\ln^2 y} = \frac{1}{(y-1)\ln y} - \frac{\ln(y-1)}{y\ln^2y}$$
$$\int_2^\infty \frac{y}{5\ln^2(y)} \frac{d(\frac{\ln(y-1)}{\ln(y)})}{dy} dy = \int_2^\infty \left(\frac{y}{5(y-1)\ln^3(y)} - \frac{\ln(y-1)}{5\ln^4(y)} \right)dy = \int_2^\infty \frac{y\ln(y) - (y-1)\ln(y-1)}{5(y-1)\ln^4(y)}dy = \int_2^\infty \frac{(y-1)(\ln(\frac{y}{y-1})) + \ln(y)}{5(y-1)\ln^4(y)}dy = \int_2^\infty\frac{\ln\left(1 +\frac{1}{y-1}\right)}{5\ln^4(y)}dy + \int_2^\infty\frac{1}{5(y-1)\ln^3(y)}dy < \int_2^\infty\frac{1.4}{5y\ln^4(y)}dy + \int_2^N\frac{1}{5(y-1)\ln^3(y)}dy + \int_N^\infty\frac{1}{5(y-1)\ln^3(y-1)}dy = \frac{1.4}{15\ln^{3}2} + \frac{1}{10\ln^{2}(N-1)} + \int_2^N\frac{1}{5(y-1)\ln^3(y)}dy$$
Since this is a fairly rough upper bound, I'm not going to bother to work it out further, but this shows it exists. (If you're wondering about the $1.4$, it's a number chosen to be slightly greater than the highest value for which $\ln(1+\frac{1}{y-1}) > y$).