Generalized exponential integral with negative argument

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Benjamin Moll
Benjamin Moll 2016-9-12
评论： Walter Roberson 2016-9-15
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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采纳的回答

Star Strider
Star Strider 2016-9-12
See if expint will do what you want. There’s also a Symbolic Math Toolbox function by the same name, so search for it if you want it instead.
The gammainc function computes the incomplete gamma function.
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Star Strider
Star Strider 2016-9-15
My (our) pleasure!
Add your vote to mine for Walter’s Answer, and Walter gets the same 4 RP’s I got.
Walter Roberson
Walter Roberson 2016-9-15
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.

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更多回答（1 个）

Walter Roberson
Walter Roberson 2016-9-12
编辑：Walter Roberson 2016-9-12
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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Benjamin Moll
Benjamin Moll 2016-9-13
Great thank you Star Strider and Walter. Both formulas work and give the same answer. One question left: I need to do this for a vector of 1000 z's, e.g. z=linspace(1,2,1000). The way you guys do it, it cannot handle vectors. Is there a way to get this to work without writing a loop?
Walter Roberson
Walter Roberson 2016-9-13
编辑：Walter Roberson 2016-9-15
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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