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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 :)

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    What do you mean "I have reviewed in several places"?2011-04-20
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    You could [differentiate under the integral sign](http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign), you know.2011-04-20
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    @J.M I think OPs real problem is evaluating either $\int e^{-ax^3-bx^2}$ or $\int x^2 e^{-ax^3-bx^2}$ over the given range. I do not think the answer will be as trivial as you indicate.2011-04-20
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    @Approximist: I see nothing about *numerical* evaluation in his post, which I'll agree is a different can of worms. Figuring out a closed form would also be interesting...2011-04-20
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    @Approximist: And upgraded the question. As I said, I have differentiated $ f(a,b) = \int_0^\infty {{e^{-a{x^3}- b{x^2}}}dx}$ under the integral, obtaining $\frac{{\partial f}}{{\partial b}} = \int_0^\infty { - {x^2}{e^{ - a{x^3} - b{x^2}}}d} x$ , but becomes more complicated at the time to show that the improper integral in the variables "a" and "b" is a solution of the differential equation: $$3ab\frac{{{\partial ^2}f}}{{\partial {b^2}}} - 3a\frac{{\partial f}}{{\partial b}} - 2{b^2}\frac{{\partial f}}{{\partial a}} = 1$$2011-04-20
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    @Theo Buehler: I tried to calculate the integral with Alpha Wolfram Web, with Maple Mathsoft software and a book of mathematical analysis of Eduardo Espinoza Ramos and found nothing similar.2011-04-20
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    So to clarify... you want a closed form?2011-04-20
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    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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    Nice work! $\;$2011-04-20
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    Nice. By the way, this is a good example that shows that it's good to include context in the question and say what one is more broadly interested in, rather than isolating the part of it that one is currently focussing on.2011-04-20
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    @J. M. Thank you very much, it became very clear reply. Now I know how to derive this type of integral functions :)2011-04-20
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    @mathsalomon: Thank Raeder, not me. I merely formatted his/her post.2011-04-20
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    @Raeder: Thank you very much, it became very clear reply. Now I know how to derive this type of integral functions :)2011-04-20
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    @mathsalomon : You're welcome! It would be great if you go edit the question so as to reflect what you were really asking :)2011-04-20
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    @Raeder Great answer! +12011-04-20
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    @Reader already upgraded the question. But I have carefully reviewed to make a correction in the calculation of the right side of the second equation, into the equation you wrote: $$= \int_0^\infty \left(3abx^4-3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx.$$ but should be written: $$= \int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx.$$ i.e. all are positive, this will cause some changes, but in essence your reasoning is fine, in fact it can prove that the expression is 1. :D2011-04-21
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    @mathsalomon: Great, I think the signs are correct now.2011-04-21