Evaluating a given limit Let $$\nu(x) = \frac{1}{2} \log{2\pi x} - x + \int\limits_{1}^{x} \frac{\lfloor{t\rfloor}}{t} \ dt$$ then what is $\displaystyle \lim_{x \to \infty} x \cdot \nu(x)$?
I really don't know to proceed. Tried something using the expansions of $\log$ and seeing how the integral behaves, but couldn't figure out. 
 A: [The answer is given at the end.]
First let $n = \left\lfloor x \right\rfloor$. We have $\int_k^{k + 1} {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t}  = k[\log (k + 1) - \log (k)]$, $k=1,\ldots,n-1$, and $\int_n^x {\frac{{\left\lfloor t \right\rfloor }}{t}\,{\rm d}t}  = n[\log (x) - \log (n)]$. From this it follows straightforwardly that 
$$
\nu(x) = \frac{1}{2}\log (2\pi x) - x - \log (n!) + n\log (x).
$$
Next, let $\delta_x = x - n$. Thus, $x = n + \delta_x$. Substituting this for $x$ in the above expression, and using 
$$
\log (M + a) = \log (M) + \frac{a}{M} - \frac{{a^2 }}{{2M^2 }} + O\bigg(\frac{1}{{M^3 }}\bigg), 
$$
as $M \to \infty$, leads to
$$
\nu (x) = \frac{1}{2}\log (2\pi n) - n - \log (n!) + n\log (n) + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),
$$
or
$$
\nu (x) = \log \Bigg[\frac{{(\sqrt {2\pi n} )e^{ - n} n^n }}{{n!}}\Bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),
$$
as $n \to \infty$. Then, from the well-known estimate
$$
n! = (\sqrt {2\pi n} )e^{ - n} n^n \bigg(1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg)
$$
(and noting that $\log(1/\xi)=-\log(\xi)$), it follows that
$$
\nu (x) =  - \log \bigg[1 + \frac{1}{{12n}} + O\bigg(\frac{1}{{n^2 }}\bigg)\bigg] + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg),
$$
and in turn,
$$
\nu (x) =  - \frac{1}{{12n}} + \frac{{\delta _x }}{{2n}} - \frac{{\delta _x^2 }}{{2n}} + O\bigg(\frac{1}{{n^2 }}\bigg).
$$
Finally, since $x = n + \delta_x$, we obtain
$$
x \nu(x) = - \frac{1}{{12}} + \frac{{\delta _x }}{{2}} - \frac{{\delta _x^2 }}{{2}} + O\bigg(\frac{1}{{n}}\bigg).$$
We conclude that $x \nu(x)$ does not have a limit as $x \to \infty$. Rather, since $\delta_x /2 - \delta_x^2 / 2$ is oscillating between $0$ and $1/8$, we have
$$
\mathop {\lim \inf }\limits_{x \to \infty } [x\nu (x)] =  - \frac{1}{{12}} \;\;\; {\rm and} \;\;\; \mathop {\lim \sup }\limits_{x \to \infty } [x\nu (x)] =  \frac{1}{{24}}.
$$
Nevertheless, we have
$$
\mathop {\lim }\limits_{n \to \infty } n\nu (n) =  - \frac{1}{{12}},
$$
as $n \to \infty$ (integer).
