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Does the logarithmic integral function $\operatorname{li}(x)$ have the continued fraction expansion $$\operatorname{li}(x) = \cfrac{x}{\log x -1 -{}} \ \cfrac{1}{\log x -3 -{}} \ \cfrac{4}{\log x -5 -{}}\ \cfrac{9}{\log x - 7-{}}\ \cfrac{16}{\log x - 9 -{}} \ \cfrac{25}{\log x - 11 -{}} \ \cdots$$ for $x > 1$? If so, is there a proof or a reference that proves it?

This question is not answered by any reference I can find because standard results in the literature will verify the identity (appropriately interpreted) for complex values of $x$ excluding $x > 1$. For $x > 1$, I do not even know if the given continued fraction converges.

It might help to note that the $n$th convergent of the continued fraction above is equal to $-\sum_{k = 1}^n \frac{1}{kL_k(x)L_{k-1}(x)}$, where $L_k(x)$ denotes the $k$th Laguerre polynomial (at least at values of $x$ that are not roots of any Laguerre polynomial). Thus, the question is equivalent to the following: does one have $\operatorname{li}(x) = -\sum_{k = 1}^\infty \frac{1}{kL_k(x)L_{k-1}(x)}$ for all real $x > 1$ that are not roots of any Laguerre polynomial?

It is well known that the exponential integral $E_1(z)$ has the continued fraction expansion $$E_1(z) = \cfrac{e^{-z}}{z+1 -{}} \ \cfrac{1}{z+3 -{}} \ \cfrac{4}{z+5 -{}}\ \cfrac{9}{z+7-{}}\ \cfrac{16}{z+9 -{}} \ \cfrac{25}{z+11 -{}} \ \cdots, \quad z \in \mathbb{C} \setminus (-\infty,0].$$ A third equivalent formulation of the question is the following: For $z \in (-\infty,0]$, does the continued fraction above converge to $-\operatorname{li}(e^{-z}) = E_1(z)+\pi i$?

EDIT: In all the references I can find, including the ones given in the proposed answer by Alexey Ustinov, the domain for which the given expansions hold exclude the domains I inquired about in my question. The domain $x> 1$ of $\operatorname{li}(x)$ is the domain number theorists care most about, and it would be nice if it had the proposed continued fraction expansion on that domain.

FINAL EDIT: I now think it's more likely that the continued fraction diverges for $x > 1$, but I don't know how to prove this.

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  • $\begingroup$ Do you really mean $\log(x) - 5 + \dotsb$ when all the others are $-$? $\endgroup$
    – LSpice
    Commented May 3, 2021 at 0:37
  • $\begingroup$ Why $+$ at $5$ exclusively? $\endgroup$
    – Turbo
    Commented May 3, 2021 at 0:38
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    $\begingroup$ typo. fixing now. $\endgroup$
    – Jesse Elliott
    Commented May 3, 2021 at 0:59

1 Answer 1

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You can find this expansion in the book Lorentzen L. & Waadeland H. Continued fractions with applications North-Holland Publishing Co., 1992 (formula (4.3.10)). As I understand, this document is a more recent version of the appendix of this book. Here the desired formula has the number (3.3.10). However, the validity of the expansion for $\operatorname{li}(x)$ does not include the domain $x > 1$ that you seek.

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    $\begingroup$ The PDF link goes to the freely available "Back matter" of the Springer (re-?)issue of the referenced book. Here's a DOI link to the book: Lorentzen and Waadeland - Continued fractions - I. $\endgroup$
    – LSpice
    Commented May 3, 2021 at 14:50
  • $\begingroup$ This won't do it. The expansion for $\operatorname{Ei}(-z)$ is proved valid only for $|\operatorname{arg}(z)| < \pi$ and the expansion for $\operatorname{li}(z)$ for $|\operatorname{arg}(-\operatorname{Ln}(z))|< \pi$, which rules out the domain of $x > 1$ for $\operatorname{li}(x)$ that I'm interested in. (I actually have a copy of the book.) $\endgroup$
    – Jesse Elliott
    Commented May 3, 2021 at 20:44
  • $\begingroup$ Again, (3.3.10) holds only for $|\operatorname{arg}(-\operatorname{Ln}(z)| < \pi$, which excludes the domain $x > 1$. Do you see the issue here? $\endgroup$
    – Jesse Elliott
    Commented May 5, 2021 at 20:50
  • $\begingroup$ I now think the continued fraction actually diverges for $x > 1$, but I don't know how to prove this. $\endgroup$
    – Jesse Elliott
    Commented May 24, 2021 at 2:26

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