upper bound for $\int \limits_{a}^{b} \frac{t \ln \ln t}{\ln t} dt$?

Question

What is the upper bound for

$$\int \limits_{a}^{b} \frac{t \ln \ln t}{\ln t} dt $$

A good approximation is sufficient but it should be of $o(\frac{1}{\ln^2 b \space or \space a})$ or something better.

When I tried to find closed form using WolframAlpha, it said that this integration has no known closed form.

Answer 1

Here's the best I could cook up:

Let $t=e^{e^u}$ and $a,b>1$ so that we have

$$I=\int_{\ln(\ln(a))}^{\ln(\ln(b))}ue^{2e^u}\ du$$

Notice then that

$$e^{2e^u}=1+2e^u+2e^{2u}+\frac43e^{3u}+\frac13e^{4u}+\mathcal O(e^{5u})$$

And by integration by parts, we find that

$$I=\frac12u^2+2(u-1)e^u+\left(u-\frac12\right)e^{2u}+\frac4{27}\left(3u-1\right)e^{3u}+\frac1{48}\left(4u-1\right)e^{4u}\bigg|_{\ln(\ln(a))}^{\ln(\ln(b))}\\+\mathcal O\left(\ln^5(b)\ln(\ln(b))\right)-\mathcal O\left(\ln^5(a)\ln(\ln(a))\right)$$

Answer 2

$$\int_{a}^{b}\frac{t^2}{2}\cdot\frac{d}{dt}\left(\log\log t\right)^2\,dt =\left.\frac{1}{2}\left(t\log\log t\right)^2\right|_{a}^{b}-\int_{a}^{b}t\left(\log\log t\right)^2\,dt$$ can be approximated very well by noticing that $\log\log t$ is approximately constant on small intervals. Without further informations on the behaviour of $a$ and $b$ this is the best I am able to give.