3
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Integrate[x^(1/3) AiryAi[x], {x, 0, ∞}]
(* Integrate[x^(1/3) AiryAi[x], {x, 0, ∞}] *)

Integrate[x^a AiryAi[x], {x, 0, ∞}] /. a -> 1/3
(* Gamma[1/3]/(3^(1/9) Gamma[1/9]) *)
  • Why do I need to generalize/parameterize an integral sometimes to get a closed form result? Should cases like this be considered as bugs?
  • How can I tell what kinds of generalization may help in a particular case? Do you know any other tricks besides replacing a constant with a parameter?
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  • 1
    $\begingroup$ What version of Mathematica are you using? In ver.8 I get the first integral unevaluated and this warning Integrate::idiv: Integral of x^(1/3) AiryAi[x] does not converge on {0, ∞}. >>. But in ver.9 the first integral evaluates to this (2^(4/9) Sqrt[Pi] Gamma[13/18] Sin[Pi/18] + Cos[Pi/18] Gamma[4/9] Gamma[7/ 9] (Cos[Pi/9] + Sqrt[3] Sin[Pi/9]))/(2 3^(7/9) Pi). This is numerically equal to the second integral. The second one behaves as you claim. $\endgroup$ – Artes Oct 20 '13 at 20:44
  • 1
    $\begingroup$ In 11.1 both return results, which agree in numerical value, but have very different forms. $\endgroup$ – Szabolcs Apr 8 '17 at 15:32

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