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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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integration by part and bound logarithmic integral. – Soarer 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. – Antonio Vargas Nov 25 at 18:22
If the default value of limits will change (from current \nolimits to \limits), this integral will looks the same. – Michael Medvinsky Nov 25 at 19:43

1 Answer 1

up vote 8 down vote accepted

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


I've given the cookie, but it's up to you to supply the cream filling...

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