Can somone help me find an assymptotic formula for n, for fixed x , for this sum , perhaps an inequality would be even better, or some bound on the error. $$\sum_{k=1}^n \frac{1}{\log(kx)}$$
I need somthing better then the integral from 1 to n of ln(kx) with respect to k. Its also okay if you use special functions, like the logarithmic integral.
The Euler-Maclaurin series for $\sum_k 1/\log(kx)$ starts $$ F(k) = \frac{\text{Li}(kx)}{x} - \frac{1}{2 \ln(kx)} - \frac{1}{12 \ln(kx)^2 k} + \frac{3 + 3 \ln(kx) + \ln(kx)^2}{360 \ln(kx)^4 k^3} + \ldots $$
That is, we should have $$\sum_{k=1}^n \frac{1}{\log(kx)} \approx F(n+1) + C$$ where $C$ is a constant.