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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 $\int\frac{e^t}{3\sqrt[3]{t^2}}dt$.

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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. – André Nicolas Jan 5 '13 at 4:40
up vote 10 down vote accepted

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).

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

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


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

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

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

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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

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This answer seems correct for $x<0$. – GEdgar May 20 '13 at 20:55

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

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

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