I seek to prove the identity


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

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

  • $\begingroup$ I think that your step from the integral to the expression $\frac{x}{\log^kx}-\frac{a}{\log^ka}$ is not correct. You need to make $a$ an increasing function of $x$ that does not increase too fast. See Captain Darling's answer for an example. $\endgroup$ – marty cohen May 14 '15 at 15:37
  • $\begingroup$ I didn't quite understand. I have chosen $a$ to be a constant, and the integral evaluates to that value. $\endgroup$ – Cody Johnson May 14 '15 at 16:01

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 $\log t$ 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.

  • $\begingroup$ Thanks, I get it now. I guess I was a little shy because I'm not a huge fan of really weak bounds, but now I know how to approach these problems. $\endgroup$ – Cody Johnson May 14 '15 at 16:01
  • 1
    $\begingroup$ Perhaps your intuition is right for this particular example. One can in fact rather easily provide an asymptotic expansion in the form $x \left(\sum_{n=0}a_n (\log x)^{-n}\right)$ for some explicit coefficients and where $a_0$ does not vanish (which is why the upper bound in the exercise is of the correct order of magnitude). $\endgroup$ – Captain Darling May 14 '15 at 19:21
  • $\begingroup$ Why do we have $\sqrt(x) = O((log x)^{-k} x)$? I know that it's true, but it's not like "oh yeah $e^x$ grows faster than $x^2$, everyone knows that". $\endgroup$ – limitIntegral314 Sep 11 '17 at 18:26

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.


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