Integral of $e^{(x-x^3)/3n}$ from $0$ to $\infty$ How can you compute the following integral assuming $n>0$?
$$\int_{x=0}^{\infty}e^{\frac{x -x^3}{3n}}dx $$
Mathematica etc. fail to produce anything useful.
EDIT: I would be happy with an asymptotic result in $n$ if it is too hard to compute exactly.  I don't know if it helps but 
$$\int_{x=0}^{\infty}e^{\frac{-x^3}{3n}}dx \approx 1.29 n^{1/3}.$$

I guess Maple has one of the useless answers:
$$\int_{0}^{\infty} \operatorname{e} ^{\frac{-x (-1 + x^{2})}{3 n}} d x = \frac{i}{9} \pi \mathrm{BesselY} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) \sqrt{3} -  \\
\quad{}\quad{}\frac{i}{9} \pi \mathrm{AngerJ} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) + \frac{2 i}{3} \mathrm{BesselK} \biggl(\frac{1}{3},\frac{-2\sqrt{3}}{27 n}\biggr) +  \\
\quad{}\quad{}\frac{i}{9} \pi \mathrm{BesselJ} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) + \frac{i}{9} \sqrt{3} \pi \mathrm{WeberE} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr)
$$
Maple reports asymptotics for this as
$$
\frac{2 \sqrt[3]{n} \pi 3^{\frac{5}{6}}}{9 \Gamma \Bigl(\frac{2}{3}\Bigr)}
$$
which differs by factor $3^{5/6}$ from Raymond's answer.  Perhaps this is more accurate??
More terms
$$
 \frac{2 \sqrt[3]{n} \pi 3^{\frac{5}{6}}}{9 \Gamma \Bigl(\frac{2}{3}\Bigr)}
+ \frac{\Gamma \Bigl(\frac{2}{3}\Bigr) 3^{\frac{2}{3}}}{9 \sqrt[3]{n}}
 - \frac{1}{36 n}
 + O(n^{-5/3})
$$
