TITLE: continued fraction for logarithmic integral
QUESTION [6 upvotes]: 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.

REPLY [4 votes]: 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.