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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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    If the default value of limits will change (from current \nolimits to \limits), this integral will looks the same.2015-11-25

1 Answers 1

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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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