Will numerical routines for the Exponential Integral function E_n work when n is continuous? So I am a mathematical biologist of sorts.  I rely heavily on Mathematica which often provides analytic results couched in terms of special functions which I then try to go and learn about.  Right now I am coming across results that include the Exponential Integral n function, E_n(x).  I am interesting in implementing these results to fit a model to data using C++.  
Having looked into E_n(x) a bit it seems that most discussions and numerical routines assume n is discrete, but the parameter Mathematica assigns to n is continuous in my application.  I'm wondering if there's any reason I should avoid treating n as a continuous variable and, if not, where I can find routines that do not assume it is discrete.  Are we safe in using open source routines and just changing the argument type? 

Your function is called the Generalized Exponential Integral  $E_p(x)$ and is described in http://dlmf.nist.gov/8.19: 
$$E_p(x) = x^{p-1}\int_x^\infty \frac{e^{-t}}{t^p}\: d t \; = \int_1^\infty \frac{e^{-xt}}{t^p}\: d t\; ,
$$
Without a specific reference to the actual open source code, I guess that 
the discrete recurrence formula based on $E_1$ or your change of variable type will not work for non-integer $p$. The continued fraction representation remains valid for the translation $n\rightarrow p\;$  and might be already used by an incomplete gamma implementation.
The easiest way to compute $E_p(x)$ for modest $p$ is the relation to the incomplete gamma function (or its normalized form $Q$) given at
http://dlmf.nist.gov/8.19.E1
$$E_p(x) = x^{p-1} \Gamma(1-p,x) = x^{p-1} Q(1-p,x)\: \Gamma(1-p)$$
The incomplete gamma function should already exist in a good numerical library for special functions (it is needed among others for statistical distributions).
For large $p$ you can use the asymptotic expansion from 
http://dlmf.nist.gov/8.20.E6 
$$ E_p(\lambda p) \sim \frac{e^{-\lambda p}}{(\lambda +1)p} \sum\limits_{k=0}^{\infty} \frac{A_k(\lambda)}{(\lambda+1)^{2k}}\;\frac{1}{p^k}$$
Note that this actually can be used for the discrete $E_n$ too.
