# Estimating $\int_e^x \log\log{t}\, dt$ so the error term in within $O\left(\frac{x}{\log^2{x}}\right)$

How can one estimate the integral $$\int_e^x \log\log{t}\, dt$$ so that the error term is within $O\left(\frac{x}{\log^2x}\right)$? We may assume that $x>e$.

Any hint?

• integration by part and bound logarithmic integral. Sep 22 '11 at 7:28
• @Michael, I'm curious, what does \nolimits do in the integral? The new version looks the same as the old to me. Nov 25 '15 at 18:22
• If the default value of limits will change (from current \nolimits to \limits), this integral will looks the same. Nov 25 '15 at 19:43

$$x\log\log\,x-\int_e^x\frac{\mathrm dt}{\log\,t}$$
The last bit evaluates to $\mathrm{li}(x)-\mathrm{Ei}(1)$, where $\mathrm{Ei}(x)$ is the exponential integral, and $\mathrm{li}(x)=\mathrm{Ei}(\log\,x)$ is the logarithmic integral.
$$x\log\log\,x-\frac{x}{\log\,x}\left(1+\frac1{\log\,x}+\frac2{(\log\,x)^2}+\cdots\right)$$