Sometimes questions like: "How many digits does 2015! have?" become quite trendy. The most reasonable approach would probably use Stirling's formula. However doing this in the same way over and over again is quite boring on thus wondered, if it were also solveable via the Euler-Maclaurin formula.
However at one point I came upon the integral: $\int_{1}^{2015} \frac{\lfloor x\rfloor}{x} dx$
And well, now I am stuck.
Does anyone know or see any way to continue from this point?
Any constructive advice, hint, comment, answer or recommendation for further reading is appreciated.
Breaking the integral in 2015 pieces: $$\begin{align} \int_{0}^{2015} \frac{\lfloor x\rfloor}{x}dx &= \sum_{k=0}^{2014} \int_{k}^{k+1} \frac{\lfloor x\rfloor}{x}dx = \sum_{k=0}^{2014} \int_{k}^{k+1} \frac{k}{x}dx = \sum_{k=1}^{2014} k \ln\frac{k+1}{k} \\ &= \sum_{k=1}^{2014} \left( (k+1) \ln (k+1) - k\ln k - \ln(k+1) \right) \\ &= \sum_{k=1}^{2014} \left( (k+1) \ln (k+1) - k\ln k \right) - \sum_{k=1}^{2014} \ln(k+1) \\ &= 2015\ln 2015 - 1\ln 1 - \sum_{k=2}^{2015} \ln k \\ &= 2015\ln 2015 - \sum_{k=2}^{2015} \ln k \\ &= 2015\ln 2015 -\ln(2015!) \end{align}$$ where to get to the fourth line we used the fact that we have a telescoping series.