How to evaluate the integral $$\int e^{x^3}dx \quad ?$$

I've tried to set $t=x^3$, but it seems to be a blind alley; I don't know what to do with $\int\frac{e^t}{3\sqrt[3]{t^2}}dt$.

  • 1
    $\begingroup$ It is very common for an elementary function not to have an elementary antiderivative. Proving this is the case for a particular function can be difficult. Your function $e^{x^3}$ happens to be one for which the standard method for showing "impossibility," which dates back in principle to Liouville, works reasonably smoothly. Many non-elementary "special functions" have been devised such that useful integrals can be expressed in terms of these special functions. I would guess that Maple, or Mathematica, even Wolfram Alpha, can produce an answer in terms of some special function. $\endgroup$ – André Nicolas Jan 5 '13 at 4:40

The antiderivative of $e^{x^3}$ cannot be expressed in terms of elementary functions. We can, however, express it using power series. Since $$ e^x = \sum_{n \geq 0} \frac{x^n}{n!}, $$ $$ e^{x^3} = \sum_{n \geq 0} \frac{(x^3)^n}{n!} = \sum_{n \geq 0} \frac{x^{3n}}{n!}.$$

You can integrate term by term to find a series representation of the antiderivative (which converges on the entire complex plane, since $e^{x^3}$ is an entire function).


$$\int e^{x^3}dx=\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx$$

$$\int \sum_{n=0}^{\infty }\frac{x^{3n}}{n!}dx=\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c$$


$$\sum_{n=0}^{\infty }\frac{x^{3n+1}}{(3n+1)(n!)}+c=x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c$$

$$x\sum_{n=0}^{\infty }\frac{(\frac{1}{3})^{(n)}(x^3)^n}{(\frac{4}{3})^{(n)}(n!)}+c=\ x\ 1F1(\frac{1}{3};\frac{4}{3};x^3)+c$$

so $$\int e^{x^3}dx=\ x\ {}_1F_1(\frac{1}{3};\frac{4}{3};x^3)+c$$

where ${}_1F_1$ is Hypergeometric Function of the First Kind


another try you can solve it with Gamma function

$$\int e^{x^3}dx=\frac{-1}{3}\int e^{-t}t^{\frac{1}{3}-1}dt$$

$$\frac{-1}{3}\int e^{-t}t^{\frac{1}{3}-1}dt=\frac{-1}{3}\int_{0}^{t}e^{-t}t^{\frac{1}{3}-1}dt+c$$

$$\frac{-1}{3}\int_{0}^{t}e^{-t}t^{\frac{1}{3}-1}dt+c=\frac{1}{3}(\int_{0}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt-\int_{t}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt)+c$$

$$\frac{-1}{3}(\int_{0}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt-\int_{t}^{\infty }e^{-t}t^{\frac{1}{3}-1}dt)+c=\frac{1}{3}\Gamma (\frac{1}{3},t)+d$$

$$\frac{1}{3}\Gamma (\frac{1}{3},t)+d=\frac{1}{3}\Gamma (\frac{1}{3},-x^3)+d$$


$$\int e^{x^3}dx=\frac{1}{3}\Gamma (\frac{1}{3},-x^3)+d$$

where d and c are constant

  • $\begingroup$ This answer seems correct for $x<0$. $\endgroup$ – GEdgar May 20 '13 at 20:55
  • $\begingroup$ @GEdgar Lack of convergence there. You'll need the lower gamma function or infinite constants of integration. $\endgroup$ – Simply Beautiful Art Feb 1 '17 at 0:48

Interestingly, the definite integral $$ \int_0^\infty e^{-x^n}dx $$ can be evaluated for any $n>0$, and is equal to $\Gamma((n+1)/n)$.

  • 2
    $\begingroup$ +1 nice, how did you come by this? $\endgroup$ – Arjang May 20 '13 at 12:41
  • $\begingroup$ Really? Doesn't it diverge for $n>0$? $\endgroup$ – Javier May 20 '13 at 18:26
  • 1
    $\begingroup$ Unless there's a minus sign in the exponential, the statement is false. $\endgroup$ – johnny May 20 '13 at 18:31
  • $\begingroup$ @johnny: thanks, of course there should be a minus sign in the exponential. $\endgroup$ – Eckhard May 20 '13 at 20:00
  • $\begingroup$ @Arjang you can prove it for example with the substitution $t=x^n$ $\endgroup$ – glS May 5 '15 at 11:05

The integral cannot be evaluated. We have to use power series of exponent and then integral term by term. $$e^{x}=\sum_{n \geq 0}{\frac{x^n}{n!}}$$


I'd like to give step by step solution to @Eckhard 's answer .

Substitude $y=x^n$ :

$\int_0^\infty e^{-x^n}dx = \frac1n\int_0^\infty e^{-y}y^{1/n-1}dy = \frac1n \Gamma(\frac 1n) = \Gamma(\frac 1n+1) = {\frac 1n}! $

$x^n=y \Rightarrow nx^{n-1}dx=dy \Rightarrow dx=\frac1nx^{1-n}dy \Rightarrow dx=\frac 1n y^{\frac 1n -1}dy$

The last integral is taken due to the main definition of gamma function .

Where :

$\Gamma(z) = \int_o^\infty e^{-x}x^{z-1}dx$

-Hope it was helpful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.