Generalized exponential integral with negative argument
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Benjamin Moll
on 12 Sep 2016
Commented: Walter Roberson
on 15 Sep 2016
Is there a way to compute the generalized exponential integral E[a,x]=int_1^infty e^(-x t) t^a dt with negative x? In particular, I need to compute the integral int_1^z Exp[-y^b/b] dy where b<0 and z>1. (I need to do this for 1000’s of z-values so numerical integration is not an option) Mathematica tells me this equals an expression involving ExpIntegralE[(-1 + b)/b, 1/b] (to be precise ExpIntegralE[(-1 + b)/b, 1/b] - z ExpIntegralE[(-1 + b)/b, z^b/b])/b. So since b<0, x=1/b is negative, and this causes a problem.
There seems to be a related thread but with no answer https://www.mathworks.com/matlabcentral/newsreader/view_thread/289192
Finally, there is an implementation of E[a,x] on file exchange https://www.mathworks.com/matlabcentral/fileexchange/52694-generalised-exponential-integral but it cannot handle negative x-values.
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Accepted Answer
Star Strider
on 12 Sep 2016
10 Comments
Star Strider
on 15 Sep 2016
My (our) pleasure!
Add your vote to mine for Walter’s Answer, and Walter gets the same 4 RP’s I got.
Walter Roberson
on 15 Sep 2016
The defining formula for Ei with two arguments is in terms of the Gamma function, which has singularities at every negative integer, so if your 1/b is a negative integer you should be expecting a singularity.
More Answers (1)
Walter Roberson
on 12 Sep 2016
Edited: Walter Roberson
on 12 Sep 2016
This formula appears to work. It requires the symbolic toolbox
(-z^(b+1)*hypergeom([(1/2)*(b+1)/b], [3/2, (1/2)*(3*b+1)/b], (1/4)*z^(2*b)/b^2)+b*z*(b+1)*hypergeom([(1/2)/b], [1/2, (1/2)*(2*b+1)/b], (1/4)*z^(2*b)/b^2)+hypergeom([(1/2)*(b+1)/b], [3/2, (1/2)*(3*b+1)/b], (1/4)/b^2)+(-b^2-b)*hypergeom([(1/2)/b], [1/2, (1/2)*(2*b+1)/b], (1/4)/b^2))/(b*(b+1))
Note: this might fail if 1/b is an integer
Reference: Maple
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Walter Roberson
on 13 Sep 2016
Edited: Walter Roberson
on 15 Sep 2016
syms b z
G = (-z^(b+1)*hypergeom([(1/2)*(b+1)/b], [3/2, (1/2)*(3*b+1)/b], (1/4)*z^(2*b)/b^2)+b*z*(b+1)*hypergeom([(1/2)/b], [1/2, (1/2)*(2*b+1)/b], (1/4)*z^(2*b)/b^2)+hypergeom([(1/2)*(b+1)/b], [3/2, (1/2)*(3*b+1)/b], (1/4)/b^2)+(-b^2-b)*hypergeom([(1/2)/b], [1/2, (1/2)*(2*b+1)/b], (1/4)/b^2))/(b*(b+1));
GG = subs(G, {z, b}, {linspace(1,2,1000), -sqrt(pi)});
GGn = double(GG);
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