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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 $$\int_{x=0}^{\infty}e^{\frac{-x^3}{3n}}dx \approx 1.29 n^{1/3}.$$

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I think numerical approximation may be you're best bet. You're going to end up with some nasty-ish infinite series to evaluate IMHO. – Alex Becker Jul 11 '12 at 21:04
up vote 5 down vote accepted

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 $e^r=1+r+\frac {r^2}2+\cdots$

$$\int_0^\infty e^{\frac{x -x^3}{3n}}dx \sim (3n)^{1/3}\left(\Gamma\left(\frac 43\right)+\frac {\Gamma\left(\frac 53\right)}{2(3n)^{2/3}}+\frac 1{2\cdot 3 (3n)^{4/3}}+\rm{O}\left(n^{-2}\right)\right)$$

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How can you rigorously show that the $e^{x}$ term has no effect on the asymptotics of this integral? – Raphael Jul 11 '12 at 21:25
Thanks very much. – Raphael Jul 12 '12 at 22:41
@Arlo: you are welcome! – Raymond Manzoni Jul 13 '12 at 9:07

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 $$ \frac{2 \sqrt[3]{n} \pi 3^{\frac{5}{6}}}{9 \Gamma \Bigl(\frac{2}{3}\Bigr)} + \frac{\Gamma \Bigl(\frac{2}{3}\Bigr) 3^{\frac{2}{3}}}{9 \sqrt[3]{n}} - \frac{1}{36 n} + O(n^{-5/3}) $$

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Yes. I can't even tell by looking it if that is a real number or not. In fact, if you do int(exp((x-x^3)/100),x=0..infinity); in Maple you get a number that is not real! I don't think it can cope with this. – Raphael Jul 11 '12 at 21:27
you are right about accuracy GEdgar (I forgot the 3 in at the bottom in one of my computations, I'll update that!) – Raymond Manzoni Jul 11 '12 at 21:42
@Arlo: If you evaluate int(exp((x-x^3)/100),x=0..infinity); to 20 places, you get $4.24 + 1.8\times 10^{-15}i$, so I think it is actually real. – GEdgar Jul 11 '12 at 21:42
@GEdgar: How do you get Maple to report the asymptotics? – Raphael Jul 11 '12 at 21:44
@Raymond: yours now agrees with mine – GEdgar Jul 11 '12 at 21:50

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