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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?

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

$$x\log\log\,x-\int_e^x\frac{\mathrm dt}{\log\,t}$$

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.

Using this asymptotic series for the exponential integral, we obtain

$$x\log\log\,x-\frac{x}{\log\,x}\left(1+\frac1{\log\,x}+\frac2{(\log\,x)^2}+\cdots\right)$$

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