$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$

Question

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

Answer 1

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

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

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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 $$\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt=I-\int\limits_{x}^{\infty}\{t\}\frac{1-3\ln{t}}{t^4}dt= I+\int\limits_{x}^{\infty}\{t\}\frac{3\ln{t}-1}{t^4}dt\leq \\ I+\int\limits_{x}^{\infty}\frac{3\ln{t}-1}{t^4}dt=I+\frac{\ln{x}}{x^3}$$

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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 $C=\frac{1+\ln{4}}{16}+I$

Answer 2

I think you refer to this formula:

Theorem. Let $f:\,\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ be a non-increasing function and such that $f\left(x\right)\rightarrow0$ as $x\rightarrow+\infty$. Then there exists some $E \in \mathbb{R}$ such that $$\sum_{n\leq x}f\left(n\right)=\int_{1}^{x}f\left(t\right)dt+E+O\left(f\left(x\right)\right)$$ as $x\rightarrow+\infty$.

So, in our case, we have: $$\sum_{n\leq x}\frac{\log\left(n\right)}{n^{3}}=\int_{1}^{x}\frac{\log\left(t\right)}{t^{3}}dt+E+O\left(\frac{\log\left(x\right)}{x^{3}}\right)$$ hence $$\sum_{n\leq x}\frac{\log\left(n\right)}{n^{3}}=-\frac{\log\left(x\right)}{2x^{2}}-\frac{1}{4x^{2}}+E+\frac{1}{4}+O\left(\frac{\log\left(x\right)}{x^{3}}\right)$$ and the formula follows by taking $C:=E+1/4$.