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I am trying to compute the following (indefinite) integral:

$$\int e^x x^{2/3} dx $$

Integrate[E^x (x)^(2/3), x]
Plot[{Re[%], Im[%]}, {x, -5, 5}]

Output is:

enter image description here

That the integral is complex for $x<0$ makes sense to me, since the integrand is. But for $x>0$ the latter is real, why is the integral complex?

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  • $\begingroup$ Your function isn't real for x<0! $\endgroup$
    – Ulrich Neumann
    Commented Jul 31, 2023 at 12:44
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    $\begingroup$ Integrate[E^x x^(2/3), {x, 0, x}] $\endgroup$
    – Domen
    Commented Jul 31, 2023 at 12:47
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    $\begingroup$ If differentiating the anti derivative gives back the integrand ($\pm$ constant in anti is OK), then the result is correct, which is in this case. !Mathematica graphics $\endgroup$
    – Nasser
    Commented Jul 31, 2023 at 12:48
  • $\begingroup$ The integrand is not real. It's multivalued (like $z^{1/3}$ is). The discontinuity arises from integrating it using a multivalued antiderivative over a path which is not analytically continuous, that is, across its branch point at the origin. Look at ComplexExpand[x^(2/3)/(-x)^(2/3)] and note how the argument of x abruptly changes from $0$ to $\pi$ across the branch point as $x$ changes from negative to positive giving rise to the discontinuity. $\endgroup$
    – josh
    Commented Jul 31, 2023 at 13:33

1 Answer 1

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The antiderivative is calculated with a complex integration constant. Its value is here

Limit[(x^(2/3)*Gamma[5/3, -x])/(-x)^(2/3), x -> 0, 
Direction -> "FromAbove"]   
(* (-(-1)^(1/3))*Gamma[5/3] *)

and you can subtract it like :

Integrate[E^x*x^(2/3), x]
Plot[{Re[% + (-1)^(1/3)*Gamma[5/3]], Im[% + (-1)^(1/3)*Gamma[5/3]], 
NIntegrate[E^y (y)^(2/3), {y, 0, x}]}, {x, 0, 1}, 
PlotStyle -> {Blue, Red, Dashed}]

enter image description here The corrected antiderivative (blue line) coincides with the numerical integration (dashed green) and the imaginary part (red) is gone.

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