A question about the constant in $\sum_{2\leqslant n\leqslant x}\frac{1}{n\log n}=\log(\log x)+B+O\left(\frac{1}{x\log x}\right)$

Question

I could prove the major part of this identity as the following: $$ \begin{align*} \sum_{2\leqslant n\leqslant x}\frac{1}{n\log n} &=\frac{1}{2\log 2}+\int_2^x \frac{1}{t\log t}dt-\int_{2}^x(t-[t])\frac{\log t+1}{t^2\log^2 t}dt-\frac{x-[x]}{x\log x}\\ &=\log(\log x)+\frac{1}{2\log 2}-\log(\log 2)-\int_{2}^xI(t)dt+O\left(\frac{1}{x\log x}\right) \end{align*}$$ where $I(t)=(t-[t])\frac{\log t+1}{t^2\log^2 t}$. Since $\int_2^x I(t)dt=\int_2^\infty-\int_x^\infty$ and $$I(t)\leqslant \frac{\log t+1}{t^2\log^2 t},$$ we get (note that $x\geqslant 2$) $$\int_{x}^\infty I(t)dt\leqslant \left.\frac{1}{t\log t}dt\right|_x^\infty=O\left(\frac{1}{x\log x}\right)$$ However, I couldn't handle the integration $\int_2^\infty I(t)dt$. It is supposed to be some constant plus $O\left(\frac{1}{x\log x}\right)$, but I am unable to prove that. Usually this kind of integration is related to Euler's constant, but in this case I couldn't get rid of $\log^2 x$ without $\leqslant$ (which will not show that it is a constant and also fail to combine it with $O\left(\frac{1}{x\log x}\right)$). $$\int_{2}^\infty I(t)dt\leqslant \left.\frac{1}{t\log t}dt\right|_2^\infty=-\frac{1}{2\log 2}$$ which is not what I should do here. Any help will be appreciated. Thanks in advance.

Answer 1

Partial summation formula used:

$$\forall (a_{n})_{n \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}}, f:]1,+\infty[ \implies \mathbb{R} \text{ }\mathcal{C}^{1}, \sum_{n=2}^{x} a_{n} f(n) = A(x)f(x) - \int_{2}^{x} A(t)f'(t)dt$$ where $A(t) = \sum_{n=2}^{t} a_{n}.$

It was too horrible to write it all in TeX, so I am going to post a picture of what I have written, I use the above formula for partial summation here:

I wish it will be clear enough for you.