On Tuesday, June 3, 2014 7:02:34 AM UTC-4, Hans W Borchers wrote:
>
> In your function I cannot calculate eint(big(-3.0)), it returns
> "nan with 256 bits of precision" (never seen `nan` with so many digits).
>
Wow, it is surprising that MPFR does not support negative arguments here.
> So I assume your function computes Ei, and not the E1 exponential integral
> while, e.g., The MATLAB function expint computes E1.
>
Too bad that MPFR computes Ei and not E1.
I agree that the E_n functions are often nicer to have than Ei.
Another alternative is to use the SciPy via PyCall, but this only gives you
a double-precision version of E1:
julia> using PyCall
@pyimport scipy.special as s
s.exp1(3.0)
0.013048381094197039
