$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$ I try to prove the following equation:
for $x \geq 2$
$$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$$
To my opinion I have to use Euler's formula, but I can't get to the right equation. 
Any help appreciated

From Euler's summation formula (also page 50 here)

$$\sum\limits_{y< n\leq x}f(n)= \int\limits_{y}^{x}f(t)dt + \int\limits_{y}^{x}\{t\}f'(t)dt-\{x\}f(x)+\{y\}f(y) \tag{1}$$

and

$$\int \frac{\ln{x}}{x^3} dx = -\frac{\ln{x}}{2x^2}- \frac{1}{4 x^2} + C \tag{2}$$

we have
$$\sum\limits_{2\leq n \leq x}\frac{\ln{n}}{n^3}=
\int\limits_{2}^{x}\frac{\ln{t}}{t^3}dt + \int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt-\{x\}\frac{\ln{x}}{x^3}+\{2\}\frac{\ln{2}}{2^3}=\\
-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}-\left(-\frac{\ln{2}}{2\cdot2^2}-\frac{1}{4\cdot2^2}\right)+\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt-\{x\}\frac{\ln{x}}{x^3}=\\
\frac{1+\ln{4}}{16}-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}+\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt-\{x\}\frac{\ln{x}}{x^3}=... \tag{3}$$

Now, $\forall x\geq2$
$$\left|\{x\}\frac{\ln{x}}{x^3}\right|\leq \frac{\ln{x}}{x^3}$$
and 
$$\left|\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt\right| \leq \int\limits_{2}^{x}\left|\{t\}\frac{1-3\ln{t}}{t^4}\right|dt \leq
\int\limits_{2}^{x}\left|\frac{1-3\ln{t}}{t^4}\right|dt=\\
\int\limits_{2}^{x}\frac{3\ln{t}-1}{t^4}dt=-\frac{\ln{x}}{x^3}+\frac{\ln{2}}{2^3}$$
this shows that 
$\int\limits_{2}^{\infty}\{t\}\frac{1-3\ln{t}}{t^4}dt=I$ exists (by taking $\lim\limits_{x\rightarrow\infty}$) and
$$\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt=I-\int\limits_{x}^{\infty}\{t\}\frac{1-3\ln{t}}{t^4}dt=
I+\int\limits_{x}^{\infty}\{t\}\frac{3\ln{t}-1}{t^4}dt\leq \\
I+\int\limits_{x}^{\infty}\frac{3\ln{t}-1}{t^4}dt=I+\frac{\ln{x}}{x^3}$$

Summarising all these together (plus definitions) and continuing from $(3)$
$$...=\frac{1+\ln{4}}{16}-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}+I+O\left(\frac{\ln{x}}{x^3}\right)+O\left(\frac{\ln{x}}{x^3}\right)=\\
\left(\frac{1+\ln{4}}{16}+I\right)-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}+O\left(\frac{\ln{x}}{x^3}\right)$$
where $C=\frac{1+\ln{4}}{16}+I$
