# Evaluating $\int_0^\infty e^{-x^n}\,\mathrm{d}x$

Is there a general approach to evaluating definite integrals of the form $\int_0^\infty e^{-x^n}\,\mathrm{d}x$ for arbitrary $n\in\mathbb{Z}$? I imagine these lack analytic solutions, so some sort of approximation is presumably required. Any pointers are welcome.

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Using the substitution $t:=x^n$ we get $\int_0^{+\infty}e^{-x^n}dx=\frac 1n\Gamma(n^{-1})$. For $n<0$ we have to look at the convergence of the integral. – Davide Giraudo Feb 18 '12 at 9:45
@Davide: maybe write an answer with slightly more detail? OP: I'll just note that $n$ a positive integer is the only interesting case, as it can be easily shown that the other cases are divergent... – J. M. Feb 18 '12 at 9:49
@Davide: Thank you very much -- that solution is much simpler than I would have thought. @ J.M.: I see that for integer $n<0$, the gamma function diverges. Thank you. – user001 Feb 18 '12 at 9:58
@J.M.: To clarify: when you say "n a positive integer is the only interesting case," you must be restricting to the integers because the OP did so. (Taken out of context, this statement is false.) – Ben Crowell Feb 18 '12 at 15:15
Yes @Ben. Of course the integral generalizes to complex numbers with positive real part... – J. M. Feb 18 '12 at 15:18

## 1 Answer

For $n=0$ the integral is divergent and if $n<0$ $\lim_{n\to\infty}e^{-x^n}=1$ so the integral is not convergent.

For $n>0$ we make the substitution $t:=x^n$, then $$I_n:=\int_0^{+\infty}e^{-x^n}dx=\int_0^{+\infty}e^{—t}t^{\frac 1n-1}\frac 1ndt=\frac 1n\Gamma\left(\frac 1n\right),$$ where $\Gamma(\cdot)$ is the usual Gamma function.

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