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I have an expression that evaluates to an expression containing multiple ExpIntegralEi expressions. However, I would prefer that Mathematica use ExpIntegralE instead. Is this possible somehow?

(Usually $\operatorname{Ei}(-x) = -E_1(x)$. However, just switching the signs does not work -- I think this relationship is not generally valid for arbitrary complex arguments).

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  • $\begingroup$ Apart, perhaps, from branch cuts, it should work, according to Wolfram MathWorld. $\endgroup$
    – bbgodfrey
    Commented Dec 8, 2015 at 23:45

1 Answer 1

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The two functions can be related with the appropriate choice of analytical extension at the branch cut. Replace

ExpIntegralEi[x + I y]

by

-ExpIntegralE[1, -x - I y] + Piecewise[{{-I Pi, y < 0}, {I Pi, y > 0}}]

A numerical comparison of the two shows that they are equal.

Plot3D[Evaluate[ReIm[ExpIntegralEi[x + I y]]], {x, -2, 2}, {y, -2, 2}, 
    AxesLabel -> {x, y, Ei}]

enter image description here

Plot3D[Evaluate[ReIm[-ExpIntegralE[1, -x - I y] + 
    Piecewise[{{-I Pi, y < 0}, {I Pi, y > 0}}]]], {x, -2, 2}, {y, -2, 2},
    AxesLabel -> {x, y, E1}]

enter image description here

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