I need help with evaluating the integral:

$$\int x e^{-x^3}dx$$


  • $\begingroup$ What do you call to solve an integral? $\endgroup$
    – Did
    Commented Apr 13, 2012 at 10:18
  • $\begingroup$ it could not be solved using elementary functions $\endgroup$ Commented Apr 13, 2012 at 10:29
  • 1
    $\begingroup$ You could relate it to an incomplete Gamma-function. And this function has been well studied. $\endgroup$ Commented Apr 13, 2012 at 10:37
  • 1
    $\begingroup$ If it were a definite integral, it could be accurately approximated. If it were $\int x^2e^{-x^3}dx$, the antiderivative can be expressed in terms of elementary functions. But for this one, it cannot. See this post covering a relevant theorem of Liouville and the Risch algorithm for more info. If this is a homework problem, it is an error. What is the precise problem or your real need? $\endgroup$
    – bgins
    Commented Apr 13, 2012 at 10:58

2 Answers 2


Typo perhaps? If it is meant to be either $\int x^2 e^{x^{3}}dx$ or $\int x e^{x^{2}} dx$ then these can be evaluated very simply.

  • $\begingroup$ Yes, it was typo in the homework. It was supposed to be $x^2 e^{x^3}dx$. Thanks anyway. $\endgroup$
    – aldo
    Commented Apr 13, 2012 at 11:28
  • $\begingroup$ No problem, glad to be of help. I take it you're good from now on? It should be obvious since I assume the very integral in question is being evaluated as you had most probably been studying "integration by inspection". $\endgroup$
    – Autolatry
    Commented Apr 13, 2012 at 11:36
  • $\begingroup$ @aldo: You might want to edit your question to account for this new development... $\endgroup$ Commented Apr 15, 2012 at 16:39
Take t = x^3 
dt = 3x^2 dx and x = t^(1/3)

and substitute x and find integration by using LIATE

  • $\begingroup$ We get $(1/3)\int e^{-t} t^{-1/3}\,dt$, unfortunately still not an "elementary" integral. $\endgroup$
    – GEdgar
    Commented Apr 13, 2012 at 13:05
  • $\begingroup$ Its look like ∫f(x)g(x)dx. $\endgroup$
    – Prasad G
    Commented Apr 13, 2012 at 13:35
  • $\begingroup$ Its look like ∫f(t)g(t)dt. Take f(t)= e^(-t) and g(t) = t^(-1/3) $\endgroup$
    – Prasad G
    Commented Apr 13, 2012 at 13:47
  • $\begingroup$ Doesn't help... $\endgroup$
    – GEdgar
    Commented Apr 13, 2012 at 16:35

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