Inverse function of $\operatorname{li}(x)$ over $x>\mu$?

How can I get the inverse function of $\operatorname{li}(x)$ over $x>\mu$?

Where $$\operatorname{li}(x)=\int_{0}^{x}\frac{ds}{\ln(s)}$$ is the so-called logarithmic integral, and $\operatorname{li}(\mu)=0$.

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$\mu$ here is the Ramanujan-Soldner constant. –  Ｊ. Ｍ. Dec 29 '11 at 3:51
$\mathrm{li}(z)=\mathrm{Ei}(\ln\,z)$; your problem here is computing the inverse of the exponential integral. –  Ｊ. Ｍ. Dec 29 '11 at 3:52
Thanks but I don't understand clearly. How can I compute the inverse of the exponential integral? is it some numerical way? –  chimpanzee Dec 29 '11 at 4:18
That's the problem. I don't see an easy way to derive a nice approximation for the exponential integral's inverse. –  Ｊ. Ｍ. Dec 29 '11 at 4:25

There are two answers. There is such an inverse function, and it is real analytic. However, as J. M. indicates, there is no evidence, in the usual places, that anyone has found an attractive asymptotic expansion for the inverse of the exponential integral function. There is a very careful treatment of this in PECINA

Just to include one item I like, for $x > 1,$ from 5.1.10 in Abramowitz and Stegun, we have $$\mbox{li} \; x = \gamma + \log \log x + \sum_{n=1}^\infty \; \frac{(\log x)^n}{n \, n!}$$ where $\gamma = 0.5772156649...$ is the Euler-Mascheroni constant.

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