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 
$$\int_{x=0}^{\infty}e^{\frac{-x^3}{3n}}dx \approx 1.29 n^{1/3}.$$
 A: The asymptotic behavior ($n\gg 1$) is given by (the $x$ contribution disappears and the integral becomes solvable) :
$$\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\Gamma\left(\frac 43\right)$$
To get the next terms of the expansion, and give you some confidence in this result, let's expand the $e^{x/(3n)}$ factor in series and use (since this integral may be rewritten as a gamma integral) :
$$\int_0^\infty \left(\frac x{3n}\right)^k e^{\frac{-x^3}{3n}}dx=\frac {(3n)^{(1-2k)/3}}{k+1}\Gamma\left(\frac{k+4}3\right)$$
Observe that we will have to divide by $(3n)^{2/3}$ at each step $k\to k+1$ (as you may see in GEdgar's answer) getting with the expansion $e^r=1+r+\frac {r^2}2+\cdots$
$$\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\left(\Gamma\left(\frac 43\right)+\frac {\Gamma\left(\frac 53\right)}{2(3n)^{2/3}}+\frac 1{2\cdot 3 (3n)^{4/3}}+\rm{O}\left(n^{-2}\right)\right)$$
A: 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
$$
 \frac{2 \sqrt[3]{n} \pi 3^{\frac{5}{6}}}{9 \Gamma \Bigl(\frac{2}{3}\Bigr)}
+ \frac{\Gamma \Bigl(\frac{2}{3}\Bigr) 3^{\frac{2}{3}}}{9 \sqrt[3]{n}}
 - \frac{1}{36 n}
 + O(n^{-5/3})
$$
