$\int_2^x\frac{dt}{\log^kt}=O\left(\frac{x}{\log^kx}\right)$

Question

I seek to prove the identity

$$\int_2^x\frac{dt}{\log^kt}=O\left(\frac{x}{\log^kx}\right)$$

I was given the following hint:

Split the integral into $\int_2^{f(x)}+\int_{f(x)}^x$ for a well-chosen function $f(x)$ with $2\le f(x)<x$ and estimate both parts from above.

but my proof was different. Can anyone (i) confirm if my proof is correct or incorrect and (ii) find the proof using the author's hint? My proof:

Pick any $a>e^k$. Then $\int_2^a\frac{dt}{\log^kt}$ is constant and finite, so it suffices to prove that $\int_a^x\frac{dt}{\log^kt}=O\left(\frac{x}{\log^kx}\right)$, which follows from

$$\left(1-\frac{k}{\log a}\right)\int_a^x\frac{dt}{\log^kt}\le\int_a^x\frac{dt}{\log^kt}\left(1-\frac{k}{\log t}\right)=\frac{x}{\log^kx}-\frac{a}{\log^ka}\le\frac{x}{\log^kx}$$

Answer 1

Here is how to use the hint:Take $f(x)=\sqrt{x}$. Then $$\int_2^{f(x)}\frac{dt}{(\log t)^k}\leq \frac{1}{(\log 2)^k}\int_2^{f(x)}1 dt$$ due to $1/\log t$ being a decreasing function in $[2,\sqrt{x}]$. The last integral equals $\sqrt{x}-2$ thus showing that the contribution of the interval $[2,\sqrt{x}]$ is $O(\sqrt{x})$. Using monotonicity we see that the remaining part contributes $$\int_{f(x)}^x\frac{dt}{(\log t)^k}\leq \frac{1}{(\log f(x))^k}\int_{f(x)}^x1 dt=\frac{1}{(\log \sqrt{x})^k}(x-\sqrt{x}),$$ which is $O((\log x)^{-k}x)$. This supersedes $\sqrt{x}$ hence the first interval makes a negligible contribution compared to the second, thus finishing your computation.

Answer 2

Another approach. You may replace $t$ with $e^u$, then you have to bound: $$ I(k)=\int_{\log 2}^{\log x}\frac{e^u}{u^k}\,du. \tag{1}$$ If we take $f_k(u)=\frac{e^k}{u^k}$, it happens that: $$ \frac{df_k}{du}=\frac{e^u}{u^{k+1}}(u-k),\qquad \frac{d^2 f_k}{du^2}=\frac{e^u}{u^{k+2}}\left(k+(u-k)^2\right),\tag{2}$$ hence $f_k$ is a positive and convex function over $\mathbb{R}^+$, with a minimum in $x=k$.

Assuming $x>e^{3k}$, it follows that: $$\begin{eqnarray*} I(k) &=& \int_{0}^{k}f_k(u)\,du + \int_{k}^{\log x}\frac{e^u}{u^k}\,du\leq C_k+\int_{k}^{\log x}\exp\left(u-k\log u\right)\,du\\ &\leq&C_k+\frac{x}{\log^k x}\int_{0}^{\log x-k}\frac{dv}{e^v(1-\frac{v}{\log x})^k}\,dv\\&\leq& C_k+\frac{x}{\log^k x}\int_{0}^{+\infty}\exp\left(-v\left(1-\frac{\frac{k}{\log x}}{1-\frac{k}{\log x}}\right)\right)\,dv\\&\leq&C_k+\frac{2x}{\log^k x},\tag{3}\end{eqnarray*}$$ a more explicit bound that substantially follows from Laplace's method.