Asymptotic of a sum evaluation as $ x \to \infty $ Let be the sum
$$ \sum_{n\le x}[x/n]=g(x) $$
where $ [x] $ means floor function.
My best try for asymptotic is $ g(x) \sim x\log (x)+\gamma x +1$
where I have used the asymptotic $ [x] \sim x $ and $ \sum_{n\le x}1/n=\gamma +\log(x)+1/x$.
Is there a way to improve this result ? Thanks.
 A: ADDED LATER:
Everything I originally wrote below was correct. However there is a much better way to look at this that I neglected to mention.
The most important feature of this sum is that it is equal to divisor summatory function $M_\tau$,
$$\sum_{n \leq x} \left[\frac{x}{n}\right] = \sum_{n \leq x} \sum_{d \leq x/n} 1 = \sum_{dn \leq x} 1 = \sum_{m \leq x} \sum_{d | m} 1 = \sum_{m \leq x} \tau(m) = M_\tau(x),$$
where $\tau(m) = \sum_{d|m}1$ is the number of divisors of $m$.
Using the hyperbola method, we can easily get the bound
$$\sum_{n \leq x} \left[\frac{x}{n}\right] = M_\tau(x) = x \log x + (2\gamma-1)x + O(x^{1/2}).$$
Of course, the divisor summatory function has been most intensively studied and better bounds are available.  The best bound to date is by Huxley, who proved
$$M_\tau(x) = x \log x + (2\gamma-1)x + O(x^{131/416 + \epsilon}).$$
The smallest $\theta$ for which the error term $O(x^{\theta+\epsilon})$ holds true is known as the Dirichlet divisor problem.  Thus we know $\theta \leq 131/416$.  But also Hardy and Landau showed that $\theta \geq 1/4$, which tells us that we should abandon our hope of ever having a constant term in our asymptotic expansion; there is no possible constant for which such an asymptotic can be true.

ORIGINAL ANSWER:
So what you write is not quite correct.
Let $\{x\} = x - [x]$ denote the fractional part of $x$.
Then $\{x\}=O(1)$ and $[x]=x+O(1)$.  So we have a crude estimate,
$$\sum_{n \leq x} \left[\frac{x}{n}\right] = \sum_{n \leq x} \left( \frac{x}{n} - \left\{\frac{x}{n}\right\} \right) = x\sum_{n \leq x} \frac{1}{n} + O\left(\sum_{n \leq x} 1 \right) = x\left(\log x + \gamma + \frac{1}{2x} + O(x^{-2})\right) + O(x) = x\log x + O(x),$$
so we have the asymptotic,

$$\sum_{n \leq x} \left[\frac{x}{n}\right] \sim x \log x.$$

This is the most that you can conclude without a more precise estimate of the sums of the fractional parts $\sum_{n \leq x}\{x/n\}$.
To attempt to get a more precise estimate, one interesting question to ask is what is
$$\lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} \left\{\frac{x}{n}\right\} = \; ?$$
If we were to assume that the fractional parts were uniformly distributed, we would conclude that that the limit would be $1/2$.  But that would be wrong!  Somewhat surprisingly, that sum is in fact a Riemann sum, and can be written as an integral,
$$\lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} \left\{\frac{x}{n}\right\} = \int_0^1 \left\{\frac{1}{t}\right\} \, dt.$$
The integrand is piecewise continuous, and we can break the integral into pieces,
$$\int_0^1 \left\{\frac{1}{t}\right\} \, dt = \sum_{k=1}^\infty \int_{1/(k+1)}^{1/k} \left(\frac{1}{t} - k\right) \, dt = \sum_{k=1}^\infty \left( \log(k+1) - \log(k) - \frac{1}{k+1}\right) = \lim_{K \rightarrow \infty} \left( \log(K+1) - \sum_{k=1}^K \frac{1}{k+1} \right) = 1 - \gamma.$$
We conclude that
$$\sum_{n \leq x} \left\{\frac{x}{n}\right\} \sim (1 - \gamma)x.$$
We can use this to refine our earlier estimate,
$$\sum_{n \leq x} \left[\frac{x}{n}\right] = \sum_{n \leq x} \frac{x}{n} - \sum_{n \leq x} \left\{\frac{x}{n}\right\} = x\left(\log x + \gamma + \frac{1}{2x} + O(x^{-2})\right) - \left(1-\gamma+o(1)\right)(x) \\ = x\log x + (2\gamma -1)x + o(x).$$
Thus, we now have a more precise asymptotic,

$$\sum_{n \leq x} \left[\frac{x}{n}\right] \sim  x\log x + (2\gamma -1)x.$$

Note we haven't justified a particular constant term in our asymptotic expansion.
