Is there any formula to calculate $ \sum_{n=2}^{x} \frac{n}{\ln(n)} $? Is there any formula to calculate $ \sum_{n=2}^{x} \frac{n}{\ln(n)} $?
As we know, there are a formula to calculate $ \sum_{n=1}^{x} \frac{\ln(n)}{n^s} $
Using the Euler-Maclaurin Sum Formula, that is mentioned on this question ( math.stackexchange)
Now, I want to know is there any formula for this summation?
 A: An approximation with error term may be derived as follows. It is not difficult to show via the Euler–Maclaurin formula that
$$
\sum\limits_{k = 2}^n {\frac{k}{{\log k}}}  = \int_2^n {\frac{t}{{\log t}}\mathrm{d}t}  + \frac{n}{{2\log n}} + K + \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right),
$$
where
$$
K = \frac{{11}}{{12\log 2}} + \frac{1}{{12\log ^2 2}} + \int_2^{ + \infty } {\left( {\left\{ t \right\}^2  - \left\{ t \right\} + \frac{1}{6}} \right)\left( {\frac{1}{{2t\log ^2 t}} - \frac{1}{{t\log ^3 t}}} \right)\!\mathrm{d}t} .
$$
In terms of the exponential integral $\operatorname{Ei}$, we can write
$$
\int_2^n {\frac{t}{{\log t}}\mathrm{d}t}  = \operatorname{Ei}(2\log n) - \operatorname{Ei}(2\log 2).
$$
Thus
$$
\sum\limits_{k = 2}^n {\frac{k}{{\log k}}}  = \operatorname{Ei}(2\log n) + \frac{n}{{2\log n}} + C + \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right),
$$
with
$$
C = K - \operatorname{Ei}(2\log 2) =  - 1.47166722580 \ldots \,.
$$
A further simplification is possible by using Theorem $8.1$ of this paper which implies
$$
\operatorname{Ei}(2\log n) = \frac{{n^2 }}{{2\log n}}\sum\limits_{m = 0}^{\left\lfloor {2\log n} \right\rfloor } {\frac{{m!}}{{(2\log n)^m }}}  + \mathcal{O}\!\left( {\frac{1}{{\sqrt {\log n} }}} \right).
$$
Accordingly,
$$
\sum\limits_{k = 2}^n {\frac{k}{{\log k}}}  = \frac{{n^2 }}{{2\log n}}\left( {\frac{1}{n} + \sum\limits_{m = 0}^{\left\lfloor {2\log n} \right\rfloor } {\frac{{m!}}{{(2\log n)^m }}} } \right) + C + \mathcal{O}\!\left( {\frac{1}{{\sqrt {\log n} }}} \right)
$$
as $n\to +\infty$.
A: So it can be concluded that there are not any exact formulae, but the best estimate for this summation is:
$$\sum_{n<x} \frac{n}{\ln{(n)}} \sim \frac{x^2}{2\ln{x}} \left(1+ \frac{1!}{2\ln{x}} +  \frac{2!}{(2\ln{x})^2}  +  \frac{3!}{(2\ln{x})^3} + \ldots \right) $$
