For the years that have gone by!

$\begin{aligned} f(n) &=\sum_{r=2}^n \int_r^{r^2} \frac 1{x\ln x} dx & \small \color{#3D99F6} \text {Let } u= \ln x, \ dx=e^u du \\ &=\sum_{r=2}^n \int_{\ln r} ^{2 \ln r} \frac {e^u} {ue^u} du \\ &=\sum_{r=2}^n \int_{\ln r} ^{2 \ln r} \frac 1u du \\ &=\sum_{r=2}^n \ln u \big|_{\ln r} ^{2 \ln r} \\ &=\sum_{r=2}^n \ln (2 \ln r) - \ln(\ln r) \\ & = \sum_{r=2}^n \ln \left(\frac {2 \ln r}{\ln r} \right) \\ &=\sum_{r=2}^n \ln 2 \\ & = (n-1)\ln 2 \end{aligned}$

$\implies f(2016) = 2015\ln 2 \implies \alpha =\boxed {2015}$