Simplifying an expression involving log using euler's constant $\gamma$

Question

I was wondering if someone could give me a hint or solution on how to obtain the following estimate: $$ x \log^2 x - 2 x \log x + 2 x + O(\log^2 x) = x \log x (\sum_{k \leq x} 1/k) - \sum_{k \leq x} (\gamma + (\gamma + 2) \log k) + O(\log^2 x). $$ Thank you very much!

Answer 1

Using the well known asymptotic for the harmonic numbers we have $$\sum_{k\leq x}\frac{1}{k}=H_{\left\lfloor x\right\rfloor }=\log\left(x\right)+\gamma+O\left(\frac{1}{x}\right) $$ hence $$x\log\left(x\right)\sum_{k\leq x}\frac{1}{k}=x\log^{2}\left(x\right)+\gamma x\log\left(x\right)+O\left(\log\left(x\right)\right)\tag{1} $$ about the other sum we have $$\sum_{k\leq x}\left(\gamma+\left(\gamma+2\right)\log\left(k\right)\right)=\gamma x+\left(\gamma+2\right)\sum_{k\leq x}\log\left(k\right)+O\left(1\right)\tag{2} $$ so we have to estimate $\sum_{k\leq x}\log\left(k\right) $. By partial summation we have $$\sum_{k\leq x}\log\left(k\right)=\left\lfloor x\right\rfloor \log\left(x\right)-\int_{1}^{x}\frac{\left\lfloor t\right\rfloor }{t}dt=x\log\left(x\right)-x+O\left(\log\left(x\right)\right)\tag{3} $$ so using $(1),\,(2),\,(3) $ our claim follows.