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My question is:

How to prove that the function:

$f(a,b)=\int_0^\infty e^{-ax^3-bx^2}\mathrm dx$

is a solution of the differential equation:

$3ab\frac{{{\partial ^2}f}}{{\partial {b^2}}} - 3a\frac{{\partial f}}{{\partial b}} - 2{b^2}\frac{{\partial f}}{{\partial a}} = 1 ?$

I have reviewed in several sites, but still nothing. :(

Thanks in advance :)

  • 0
    As for closed forms... it's not one, but it's awfully similar to a [Scorer function](http://dlmf.nist.gov/9.12). You probably might be able to derive an expression in terms of Airy/Scorer functions, plus some hypergeometric term.2011-04-20

1 Answers 1

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If the goal of the question is to prove that $f(a,b) = \int_0^\infty e^{-ax^3-bx^2}\mathrm dx,$ satisfies $3ab\frac{\partial^2 f}{\partial b^2}-3a\frac{\partial f}{\partial b}-2b^2\frac{\partial f}{\partial a} = 1,$ then this can be done by looking at all terms at the same time, not separately.

I will assume that the differentiating under the integral sign is not the problematic part.

By plugging in, we get the following: $3ab\frac{\partial^2f}{\partial b^2}-3a\frac{\partial f}{\partial b}-2b^2\frac{\partial f}{\partial a} = \int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx.$

Now, observe that $\int_0^\infty 3ax^2e^{-ax^3 - bx^2}\mathrm dx = \int_0^\infty ax^3\left(3ax^2 + 2bx\right)e^{-ax^3-bx^2}\mathrm dx ,$ by integration by parts.

This means that, after factoring, we obtain $\int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx = \int_0^\infty \left(3ax^2+2bx\right) \left(ax^3+bx^2\right) e^{-ax^3-bx^2}\mathrm dx.$ Using the substitution $u = ax^3 + bx^2$, this reduces to $\int_0^\infty ue^{-u}\mathrm du = 1,$ whence we get the wanted identity.

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    @mathsalomon: Great, I think the signs are correct now.2011-04-21